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V 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsofanalytOOsmitrich 


THE  ELEMENTS  OF 


ANALYTIC    GEOMETRY 


BY 


PERCEY  F.  SMITH,  Ph.D. 

Professor  of  Mathematics  in  the  Sheffield  Scientific  School 
Yale  University 


ARTHUR    SULLIVAN   GALE,  Ph.D. 

Fayerweather  Professor  of  Mathematics  in 
The  University  of  Rochester 


1^ 


GINN  &  COMPANY 

BOSTON  .  NEW  YORK  •  CHICAGO  •  LONDON 


^'>^'?-°' 


SCNEUl 


Copyright,  1904,  by 
ARTHUR  SULLIVAN  GALE 


ALL  RIGHTS  RESERVED 
99.1 


^fae   iatbtnaum  l^vtest 

GINN   &   COMPANY  .  PRO- 
PRIETORS .  BOSTON  •  U.S.A. 


PREFACE 

In  preparing  this  volume  the  authors  have  endeavored  to  write 
a  drill  book  for  beginners  which  presents  the  elements  of  the 
subject  in  a  manner  conforming  with  modern  ideas.  The  scope 
of  the  book  is  limited  only  by  the  assumption  that  a  knowledge 
of  Algebra  through  quadratics  must  suffice  for  any  investigation. 
This  does  not  mean  a  treatise  on  conic  sections.  In  fact,  the 
authors  have  intentionally  avoided  giving  the  book  this  form. 
Conic  sections  naturally  appear,  but  chiefly  as  illustrative  of 
general  analytic  methods.  A  chapter  is  devoted  to  their  study, 
but  the  numerous  properties  of  these  curves  are  developed  inci- 
dentally as  applications  of  methods  of  general  importance. 

The  subject-matter  is  rather  more  than  is  necessary  for  the 
usual  course  of  sixty  exercises.  It  has  been  made  so  intentionally, 
to  permit  of  choice  on  the  part  of  the  teacher,  and  also  in  order 
to  include  all  topics  strictly  elementary  in  the  sense  defined 
above.  The  table  of  contents  will  show  topics  not  usually  treated. 
For  example,  in  discussing  the  nature  of  the  locus  of  the  general 
equation  of  the  second  degree  (Chapter  XII),  invariants  are 
introduced.  Again,  three  chapters  are  devoted  to  the  simple 
transformations  in  the  plane.  After  mastering  the  entire  book, 
the  student  is  assured  of  an  acquaintance  with  all  that  is  funda- 
mental in  modern  Analytic  Euclidean  Geometry. 

Attention  is  called  to  the  method  of  treatment.  The  subject  is 
developed  after  the  Euclidean  method  of  definition  and  theorem. 


196499 


iv  PREFACE 

without,  however,  adhering  to  formal  presentation.  The  advan- 
tage is  obvious,  for  the  student  is  made  sure  of  the  exact  nature 
of  each  acquisition.  Again,  each  method  is  summarized  in  a  rule 
stated  in  consecutive  steps.  This  is  a  gain  in  clearness.  Many 
illustrative  examples  are  worked  out  in  the  text. 

Emphasis  has  everywhere  been  put  upon  the  analytic  side, 
that  is,  the  student  is  taught  to  start  from  the  equation.  He  is 
shown  how  to  work  with  the  figure  as  a  guide,  but  is  warned  not 
to  use  it  in  any  other  way.  Chapter  III  may  be  referred  to  in 
this  connection. 

The  same  methods  have  been  used  uniformly  for  the  plane  and 
for  space.  In  this  way  the  extension  to  three  dimensions  is  made 
easy  and  profitable. 

Acknowledgments  are  due  to  Dr.  W.  A.  Granville  for  many 
helpful  suggestions,  to  Professor  E.  H.  Lockwood  for  suggestions 
regarding  some  of  the  drawings,  and  to  Mr.  L.  C.  Weeks  for 
assistance  in  proof  reading. 

New  Haven,  Connecticut 
December,  1904 


CONTENTS 


CHAPTER  I 
REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 

SECTION  PAGE 

1.  Numbers 1 

2.  Constants 1 

3.  The  quadratic.    Typical  form 2 

4.  Special  quadratics 4 

5.  Cases  when  the  roots  of  a  quadratic  are  not  independent  .         .6 

6.  Variables 10 

7.  Variation  in  sign  of  a  quadratic 10 

8.  Infinite  roots .         .         .  14 

9.  Equations  in  several  variables 16 

10.  Functions  of  an  angle  in  a  right  triangle 18 

11.  Angles  in  general 18 

12.  Formulas  and  theorems  from  Trigonometry  ....         19 

13.  Natural  values  of  trigonometric  functions         . '       .         .         .         .21 

14.  Rules  for  signs 22 

15.  Greek  alphabet  .         . 22 


CHAPTER  II 


CARTESIAN  COORDINATES 


16.  Directed  line 

17.  Cartesian  coordinates 

18.  Rectangular  coordinates 

19.  Angles       .... 

20.  Orthogonal  projection   . 

21.  Lengths     . 

22.  Inclination  and  slope    . 

23.  Point  of  division 

24.  Areas    .... 

25.  Second  theorem  of  projection 


23 
24 
25 
28 
29 
31 
34 
38 
42 
47 


Vi  CONTENTS 


CHAPTER  III 
THE  CURVE  AND  THE  EQUATION 

SECTION  PAGE 

26.  Locus  of  a  point  satisfying  a  given  condition 51 

27.  Equation  of  the  locus  of  a  point  satisfying  a  given  condition       .  61 

28.  First  fundamental  problem 53 

29.  General  equations  of  the  straight  line  and  circle          ...  57 

30.  Locus  of  an  equation 59 

31.  Second  fundamental  problem 60 

32.  Principle  of  comparison 62 

33.  Third  fundamental  problem.    Discussion  of  an  equation    .         .  67 

34.  Symmetry .         .72 

35.  Further  discussion .  73 

36.  Directions  for  discussing  an  equation 74 

37.  Points  of  intersection 76 

38.  Transcendental  curves ^        .         .79 

39.  Graphical  representation  in  general .  83 


CHAPTER  IV 
THE  STRAIGHT  LINE  AND  THE  GENERAL  EQUATION  OF  THE  FIRST  DEGREE 

40.  Introduction .         .         .85 

41.  The  degree  of  the  equation  of  a  straight  line       ....  85 

42.  The  general  equation  of  the  first  degree,  Ax  +  By  -]-  C  =  0    .         .  86 

43.  Geometric  interpretation  of  the  solution  of  two  equations  of  the 

first  degree 89 

44.  Straight  lines  determined  by  two  conditions 92 

45.  The  equation  of  the  straight  line  in  terms  of  its  slope  and  the  coordi- 

nates of  any  point  on  the  line 95 

46.  The  equation  of  the  straight  line  in  terms  of  its  intercepts      .         .  96 

47.  The  equation  of  the  straight  line  passing  through  two  given  points  97 

48.  The  normal  form  of  the  equation  of  the  straight  line      .         .         ,  101 

49.  The  distance  from  a  line  to  a  point 105 

50.  The  angle  which  a  line  makes  with  a  second  line    ....  109 

51.  Systems  of  straight  lines 113 

52.  The  system  of  lines  parallel  to  a  given  line li6 

53.  The  system  of  lines  perpendicular  to  a  given  line        .        .        .  117 

54.  The  system  of  lines  passing  through  the  intersection  of  two  given 

lines     . 119 

55.  The  parametric  equations  of  the  straight  line      ....  123 


CONTENTS  vii 


CHAPTER  V 

THE  CIRCLE  AND  THE  EQUATION  ac^  +  y"- +  I)x  +  Ey  +  F  =  0 

SECTION  PAGE 

56.  The  general  equation  of  the  circle 130 

57.  Circles  determined  by  three  conditions 132 

58.  Systems  of  circles 136 

59.  The  length  of  the  tangent 144 


CHAPTER  VI 

POLAR  COORDINATES 

60.  Polar  coordinates 149 

61.  Locus  of  an  equation 150 

62.  Transformation  from  rectangular  to  polar  coordinates    .         .         .  154 

63.  Applications 156 

64.  Equation  of  a  locus •  157 

CHAPTER  VII 

TRANSFORMATION  OF  COORDINATES 

65.  Introduction 160 

66.  Translation  of  the  axes 160 

67.  Rotation  of  the  axes ,        .  162 

68.  General  transformation  of  coordinates 163 

69.  Classification  of  loci '  164 

70.  Simplification  of  equations  by  transformation  of  coordinates  .  165 

71.  Application  to  equations  of  the  first  and  second  degrees      .         .  168 

CHAPTER  VIII 

CONIC  SECTIONS  AND  EQUATIONS  OF  THE  SECOND  DEGREE 

72.  Equation  in  polar  coordinates 173 

73.  Transformation  to  rectangular  coordinates  ....  178 

74.  Simplification  and  discussion  of  the  equation  in  rectangular  coordi- 

nates.   The  parabola,  e  =  \ 178 

75.  Simplification  and  discussion  of  the  equation  in  rectangular  coordi- 

nates.   Central  conies,  e  ^  1 182 

76.  Conjugate  hyperbolas  and  asymptotes -.  189 

77.  The  equilateral  hyperbola  referred  to  its  asymptotes  .         .         .  191 


viii  CONTENTS 

SECTION  PAGE 

78.  Focal  property  of  central  conies 192 

79.  Mechanical  construction  of  conies 192 

80.  Types  of  loci  of  equations  of  the  second  degree     .         .         .        .194 

81.  Construction  of  the  locus  of  an  equation  of  the  second  degree  .         197 

82.  Systems  of  conies 200 

CHAPTER  IX 

TANGENTS  AND  NORMALS 

83.  The  slope  of  the  tangent 207 

84.  Equations  of  tangent  and  normal 210 

85.  Equations  of  tangents  and  normals  to  the  conic  sections   .         .  212 

86.  Tangents  to  a  curve  from  a  point  not  on  the  curve        .         .  .     215 

87.  Properties  of  tangents  and  normals  to  conies     .         .         .         .  217 

88.  Tangent  to  a  curve  at  the  origin    .         .         .         .         .         .  .221 

89.  Second  method  of  finding  the  equation  of  a  tangent          .         .  223 


CHAPTER  X 

RELATIONS  BETWEEN  A  LINE  AND  A  CONIC.     APPLICATIONS  OF  THE  THEORY 

OF  QUADRATICS 

90.  Relative  positions  of  a  line  and  conic 226 

91.  Relative  positions  of  lines  of  a  system  and  a  conic,  and  of  a  line  and 

conies  of  a  system 228 

92.  Tangents  to  a  conic 2.30 

93.  Tangent  in  terms  of  its  slope 233 

94.  The  equation  in  p 235 

95.  Tangents 237 

96.  Asymptotic  directions  and  asymptotes  .  • 238 

97.  Centers 240 

98.  Diameters 241 

99.  Conjugate  diameters  of  central  conies 244 

CHAPTER  XI 

LOCI.    PARAMETRIC  EQUATIONS 

100.  Introduction 248 

101.  Loci  defined  by  a  construction  and  a  given  curve      .        .         .  249 

102.  Parametric  equations  of  a  curve 253 

103.  Loci  defined  by  the  points  of  intersection  of  systems  of  curves  .  259 


CONTENTS  ix 


CHAPTER  XII 
THE  GENERAL  EQUATION  OF  THE  SECOND  DEGREE 

SECTION  PAGE 

104.  Introduction 204 

105.  Condition  for  a  degenerate  conic 264 

106.  Degenerate  conies  of  a  system 267 

107.  Invariants  under  a  rotation  of  the  axes 269 

108.  Invariants  under  a  translation  of  the  axes 273 

109.  Nature  of  the  locus  of  an  equation  of  the  second  degree    .         .  275 

110.  Equal  conies 278 

111.  Conies  determined  by  five  conditions 279 


CHAPTER  XIII 

EUCLIDEAN  TRANSFORMATIONS  WITH  AN  APPLICATION  TO  SIMILAR  CONICS 

112.  Introduction 281 

113.  Equal  figures 281 

114.  Translations 282 

115.  Rotations 282 

116.  Displacements        . 284 

117.  The  reflection  in  a  line 287 

118.  Symmetry  transformations 287 

119.  Congruent  and  symmetrical  conies     .         .         .         .         .         .  291 

120.  nomothetic  transformations 291 

121.  Similitude  transformations 292 

122.  Similar  conies 293 


CHAPTER  XIV 

INVERSION 

123.  Definition 297 

124.  Equations  of  an  inversion 297 

125.  Inversion  of  conic  sections 299 

126.  Angle  formed  by  two  circles 303 

127.  Angles  invariant  under  inversion 304 

128.  Inversion  of  systems  of  straight  lines 306 

129.  Inversion  of  a  system  of  concentric  circles         ....  307 

130.  Orthogonal  systems  of  circles 308 


xii  CONTENTS 


CHAPTER  XXII 


QUADRIC  SURFACES  AND  EQUATIONS  OF  THE  SECOND  DEGREE 
IN  THREE  VARIABLES 

SECTION  PAGE 

176.  Quadric  surfaces ' 397 

177.  Simplification  of  the  general  equation  of  the  second  degree  in 

three  variables 398 

178.  The  ellipsoid  ^  + ^  +  ^  =  1 400 

a2      62       c2 

a^2         y1         ^1 

179.  The  hyperboloid  of  one  sheet  ~  -^  ~ =  1  .         .         .         .         401 

a^      h:      c^ 

^•2  yl  ^1 

180.  The  hyperboloid  of  two  sheets =  1     .         .         .         .402 

181.  The  elliptic  paraboloid  ^  +  ^  =  202        .        .         .         .         .         405 

182.  The  hyperbolic  paraboloid ---- =r  2  cz       .         .       >         .         .     406 

183.  Rectilinear  generators 408 


CHAPTER  XXIII 

RELATIONS  BETWEEN  A  LINE  AND  QUADRIC.     APPLICATIONS  OF  THE 
THEORY  OF  QUADRATICS 

184.  The  equation  in  p.     Relative  positions  of  a  line  and  quadric          .  410 

185.  Tangent  planes 411 

186.  Polar  planes 412 

187.  Circumscribed  cones 412 

188.  Asymptotic  directions  and  cones    .......  415 

189.  Centers 419 

190.  Diametral  planes 419 

Index 423 


i 


AINTALYTIC   GEOMETRY 

CHAPTER   I 
REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 

1.  Numbers.  The  numbers  arising  in  carrying  out  the  opera- 
tions of  Algebra  are  of  two  kinds,  real  and  imaginary. 

A  real  number  is  a  number  whose  square  is  a  positive  number. 
Zero  also  is  a  real  number. 

A  pure  imaginary  number  is  a  number  whose  square  is  a  nega- 
tive number.  Every  such  number  reduces  to  the  square  root  of 
a  negative  number,  and  hence  has  the  form  6  V—  1,  where  ^  is  a 
real  number,  and  (V—  1)^  =  —  1. 

An  imaginary  or  complex  number  is  a  number  which  may  be 
written  in  the  form  a-\-b  V—  1,  where  a  and  b  are  real  numbers, 
and  b  is  not  zero.  Evidently  the  square  of  an  imaginary  number 
is  in  general  also  an  imaginary  number,  since 

(a^b  V^)2  =  a''-b''-\-2ab  V^^, 

which  is  imaginary  if  a  is  not  equal  to  zero. 

2.  Constants.  A  quantity  whose  value  remains  unchanged  is 
called  a  constant. 

Numerical  or  absolute  constants  retain  the  same  values  in  all 
problems,  as  2,  —  3,  Vl,  tt,  etc. 

Arbitrary  constants,  or  parameters,  are  constants  to  which  any 
one  of  an  unlimited  set  df  numerical  values  may  be  assigned,  and 
these  assigned  values  are  retained  throughout  the  investigation. 

Arbitrary  constants  are  denoted  by  letters,  usually  by  letters  from  the 
first  part  of  the  alphabet.     In  order  to -increase  the  number  of  symbols  at  our 

1 


2  .     ANALYTIC  GEOMETRY 

disposal,  it  is  convenient  to  use  primes  (accents)  or  subscripts  or  both.  For 
example : 

Using  primes, 

a' (read  "a  prime  or  a  first  ")j  a'' (read  "a  double  prime  or  a  second"), 
a'''(read  "a  third"),  are  all  different  constants. 

Using  subscripts, 

&i  (read  "  b  one  "),  b^  (read  "  b  two  "),  are  different  constants. 

Using  both, 

c/ (read  "c  one  prime"),  Cg'' (read  "c  three  double  prime"),  are  different 
constants. 

3.  The  quadratic.  Typical  form.  Any  quadratic  equation 
may  by  transposing  and  collecting  the  terms  be  written  in  the 
Typical  Form 


(1)  Ax'^  +  Bx  +  C  =  0, 

in  which  the  unknown  is  denoted  by  x.  The  coefficients  A,  B,  C 
are  arbitrary  constants,  and  may  have  any  values  whatever, 
except  that  A  cannot  equal  zero,  since  in  that  case  the  equation 
would  be  no  longer  of  the  second  degree.  C  is  called  the  con- 
stant term. 

The  left-hand  member 

(2)  Ax""  -\-Bx-\-C 

is  called  a  quadratic,  and  any  quadratic  may  be  written  in  this 
Typical  Form,  in  which  the  letter  x  represents  the  unknown. 
The  quantity  B^  —  AAC  is  called  the  discriminant  of  either  (1) 
or  (2),  and  is  denoted  by  A. 

That  is,  the  discriminant  A  of  a  quadratic  or  quadratic  equa- 
tion in  the  Typical  Form  is  equal  to  the  square  of  the  coefficient 
of  the  first  power  of  the  unknown  diminished  by  four  times  the 
product  of  the  coefficient  of  the  second  power  of  the  unknown 
by  the  constant  term. 

The  roots  of  a  quadratic  are  those  numbers  which  make  the 
quadratic  equal  to  zero  when  substituted  for  the  unknown. 

The  roots  of  the  quadratic  (2)  are  also  said  to  be  roots  of  the 
quadratic  equation  (1).  A  root  of  a  quadratic  equation  is  said 
t()  satisfy  that  equation. 


l-^--^l^r^^c. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  3 

In  Algebra  it  is  shown  that  (2)  or  (1)  has  two  roots,  x^  and  x^, 
obtained  by  solving  (1),  namely, 

(3) 

Adding  these  values,  we  have 

(4)  x^^-x^=--' 
Multiplying  gives 

(5)  ^  <cxx^  =  -' 

Hence 

Theorem  I.  The  sum  of  the  roots  of  a  quadratic  is  equal  to  the 
coefficient  of  the  first  power  of  the  unknown  with  its  sign  changed 
divided  by  the  coefficient  of  the  second  power. 

The  product  of  the  roots  equals  the  constant  term  divided  by  the 
coefficient  of  the  second  power. 

The  quadratic  (2)  may  be  written  in  the  form 

(6)  Ax^  -^Bx  +  C  =*^(a;  -  x^  {x  -  x^), 

as  may  be  readily  shown  by  multiplying  out  the  right-hand 
member  and  substituting  from  (4)  and  (5). 

For  example,  since  the  roots  of  3a;2  —  4a;  +  l  =  0  are  1  and  I,  we  have  iden- 
tically 3x2--  4a;  +  1  =  3  (x  -  1)  (cc  -  i). 

The  character  of  the  roots  x^  and  x^  as  numbers  (§  1)  when  the 
coefficients  A,  B,  C  are  real  numbers  evidently  depends  entirely 
upon  the  discriminant.    This  dependence  is  stated  in 

Theorem  II.  If  the  coefficients  of  a  quadratic  are  real  numbers, 
and  if  the  discriminant  be  denoted  by  A,  then 

when  A  is  positive  the  roots  are  real  and  unequal; 
when  A  is  zero  the  roots  are  real  and  equal; 
when  A  is  negative  the  roots  are  imaginary. 

*  The  sign  =  is  read  "  is  identical  with,"  and  means  that  the  two  expressions 
connected  hy  this  sign  differ  only  inform. 


4  ANALYTIC  GEOMETRY 

In  the  three  cases  distinguished  by  Theorem  II  the  quadratic 
may  be  written  in  three  forms  in  which  only  real  numbers  appear. 
These  are 

'  Ax^-\-Bx~\-C  =  A  (x—x^  (x  —  ccg),  from  (6),  if  A  is  positive ; 
Ax^  -\-  Bx  -\-C  =  A  (x  —  XiY,  from  (6),  if  A  is  zero ; 


(7) 


I  37  +  ^— I  H — ^ —  I  if  A  is  negative. 


The  last  identity  is  proved  thus : 

Ax^  +  Bx+C=a(x2-\--z-\--) 

/   „      5  B^       C       B'^\ 

B^ 
adding  and  subtracting  ——  within  the  parenthesis. 

.■.Ax^+Bx+C=A[(x  +  ^y+^-^^f^^'  Q.E.D. 

4.  special  quadratics.    If  one  or  both  of  the  coefficients  B  and 
C  in  (1),  p.  2,  is  zero,  the  quadratic  is  said  to  be  special. 

Case  I.    C  =  0. 

Equation  (1)  now  becomes,  by  factoring, 

(1)  -  Ax^  -\-Bx  =  x(Ax-\-B)=  0. 

Hence  the  roots  are  Xi  =  0,  x^  =—  —•    Therefore  one  root  of 

a  quadratic  equation  is  zero  if  the  constant  term  of  that  equation 
is  zero.  And  conversely,  if  zero  is  a  root  of  a  quadratic,  the  con- 
stant term  must  disappear.  For  if  a;  =  0  satisfies  (1),  p.  2,  by 
substitution  we  have  C  =  0. 

Case  II.   ^  =  0.  , 

Equation  (1),  p.  2,  now  becomes 

(2)  Ax^+C  =  0. 

From  Theorem  I,  p.  3,  £^1  +  3^2  =  0,  that  is, 

(3)  Xi  =  —  x^. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  5 

Therefore,  if  the  coefficient  of  the  first  power  of  the  unknown 
in  a  quadratic  equation  is  zero,  the  roots  are  equal  numerically 
but  have  opposite  signs.  Conversely,  if  the  roots  of  a  quadratic 
equation  are  numerically  equal  but  opposite  in  sign,  then  the 
coefficient  of  the  first  power  of  the  unknown  must  disappear.  For, 
since  the  sum  of  the  roots  is  zero,  we  must  hav^,  by  Theorem  I, 
B  =  0. 

Case  III.   B  =  C  =  0. 

Equation  (1),  p.  2,  now  becomes 

(4)  Ax'^  =  0. 

Hence  the  roots  are  both  equal  to  zero,  since  this  equation 
requires  that  x^  =  0,  the  coefficient  A  being,  by  hypothesis, 
always  different  from  zero. 

5.  Cases  when  the  roots  of  a  quadratic  are  not  independent. 

If  a  relation  exists  between  the  roots  Xi  and  X2  of  the  Typical 

then  this  relation  imposes  a  condition  upon  the  coefficients  A, 
B,  and  C,  which  is  expressed  by  an  equation  involving  these 
constants. 

For  example,  if  the  roots  are  equal,  that  is,  if  Xi  =  x^,  then 
B'^-4:AC  =  0,  by  Theorem  II,  p.  3. 

Again,  if  one  root  is  zero,  then  ajiCCa  =  0 :  hence  C  =  0,  by 
Theorem  I,  p.  3. 

This  correspondence  may  be  stated  in  parallel  columns  thus : 

Quadratic  in  Typical  Form 

Relation  between  the  Equation  of  condition  satisfied 

roots  hy  the  coefficients 

In  many  problems  the  coefficients  involve  one  or  more  arbitrary 
constants,  and  it  is  often  required  to  find  the  equation  of  condi- 
tion satisfied  by  the  latter  when  a  given  relation  exists  between 
the  roots.     Several  examples  of  this  kind  will  now  be  worked  out 


6  Al^IALYTIC  GEOMETRY 

Ex.  1.    What  must  be  the  value  of  the  parameter  k  if  zero  is  a  root  of 
the  equation 

(1)  2x2-6x  +  A:2-3A:-4  =  0? 

Solution.    Here  A  =  2,  B  =  - 6,  C  =  k^  -  Sk  -  ^.     By  Case  I,  p.  4,  zero 
is  a  root  when,  and  only  when,  C  =  0. 

.-.  A;2  -  3  ^  -  4  =  0. 
Solving,  k  =  4  or  —  1.     Ans. 

Ex.  2.    For  what  values  of  k  are  the  roots  of  the  equation 

kx^  +  2kx-4x^2  -3k 
real  and  equal  ? 

Solution.    Writing  the  equation  in  the  Typical  Eorm,  we  have 

(2)  kx^  +  {2k  -  4:)x  +  {Sk  -  2)  =  0. 
Hence,  in  this  case, 

A  =  k,  B  =  2k-4,  C  =  Sk-2. 
Calculating  the  discriminant  A,  we  get  ^ 

A  =  (2  fc  -  4)2  -  4  A:  (3  A;  -  2) 
=:  -  8  A;2  -  8  A:  +  16  =  -  8  (A;2  +  ^  -  2). 

By  Theorem  II,  p.  3,  the  roots  are  real  and  equal  when,  and  only  when, 

^  "  ^"  .'.  k^  +  k -2  =  0. 

Solving,  k=—2  or  1,     Ans. 

Verifying  by  substituting  these  answers  in  the  given  equation  (2) : 
when  k=-2,  the  equation  (2)  becomes  -2ic2-8x-8=0,  or  -2  (x+ 2)2=0; 
whenA;=     1,  the  equation  (2)  becomes        x2— 2x+l=0,  or        {x  — 1)^=0. 

Hence,  for  these  values  of  k,  the  left-hand  member  of  (2)  may  Ije  trans- 
formed as  in  (7),  p.  4. 

Ex.  3.    What  equation  of  condition  must  be  satisfied  by  the  constants 
a,  6,  k,  and  m  if  the  roots  of  the  equation 

(3)  (62  +  a2m2)  ?/2  +  2  a'^k7ny  +  a^  -  a^b^  =  0 
are  equal  ? 

Solution.    The  equation  (3)  is  already  in  the  Typical  Form  ;  hence 
^  =  62  _^  (j2^2^  B=2  a^km,  C  =  a^k^  -  a^b^. 

By  Theorem  II,  p.  3,  the  discriminant  A  must  vanish  ;  hence 
A  =  4  a'^k'^m^  -  4  (62  -f-  a2m2)  (a2A;2  _  a262)  =  o. 

Multiplying  out  and  reducing, 

a262  (A;2  -  a2„i2  _  52)  :^  0.     Ans. 


I 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  7 

Ex.  4,  For  what  values  of  k  do  the  common  solutions  of  the  simultaneous 
equations 

(4)  3  X  +  4  2/  =  /c, 

(5)  x2  +  2/2  _  25 
become  identical  ? 

Solution.    Solving  (4)  for  y,  we  have 

(6)  y  =  l{k-3x). 
Substituting  in  (5)  and  arranging  in  the  Typical  Form  gives 

(7)  25x2  -  6 fcx  +  A;2  -  400  =  0. 

Let  the  roots  of  (7)  be  Xi  and  x^.  Then  substituting  in  (6)  will  give  the 
corresponding  values  yi  and  ?/2  of  y,  namely, 

(8)  yi  =  l{k-  3xi),  2/2  =  i(A:  -  8x2), 

and  we  shall  have  two  common  solutions  (xi,  2/1)  and  (X2,  2/2)  of  (4)  and  (5). 
But,  by  the  condition  of  the  problem,  these  solutions  must  be  identical. 
Hence  we  must  have 

(9)  xi  =  X2  and  2/1  =  2/2- 

If,  however,  the  first  of  these  is  true  (xi  =  X2),  then  from  (8)  2/1  and  2/2 
will  also  be  equal. 

Therefore  the  two  common  solutions  of  (4)  and  (5)  become  identical  when, 
and  only  when,  the  roots  of  the  equation  (7)  are  equal;  that  is,  when  the  dis- 
criminant A  of  (7)  vanishes  (Theorem  II,  p.  3). 

...  A  =  36  A;2  -  100  {k'^  -  400)  =  0. 
Solving,  A:2  =  625, 

A;  =  25  or  -  25.     Ans. 

Verification.    Substituting  each  value  of  k  in  (7), 
when  A;  =  25,  the  equation  (7)  becomes  x2-6  x  + 9  =  0,  or  (x- 3)2  =  0  ;  .-.  x  =  3  ; 
when  A;=- 25,  the  equation  (7)  becomes  x2+6x  + 9=0,  or  (x+3)2=0;  .•.x=-3. 

Then  from  (6),  substituting  corresponding  values  of  k  and  x, 
when  k  =      25  and  x=      3,  we  have  y  =  ^{26  —  9)=  4^; 
when  A;  =  —  25  and  x  =  —  3,  we  have  y  =  ^ (—  25  4-  9)  =  —  4. 

Therefore  the  two  common  solutions  of  (4)  and  (5)  are  identical  for  each 
of  these  values  of  k,  namely, 

if  A;  =      25,  the  common  solutions  reduce  to  x  =  3,  y  =  4 ; 
if  A;  =  —  25,  the  common  solutions  reduce  tox  =  —  3,  y  =  —  4. 

Q.E.D. 


ANALYTIC  GEOMETRY 


PROBLEMS 


1.  Calculate  the  discriminant  of  each  of  the  following  quadratics,  deter- 
mine the  sum,  the  product,  and  the  character  of  the  roots,  and  write  each 
quadratic  in  one  of  the  forms  (7),  p.  4. 

(a)  2x2  -  6x  +  4.  (i)  5x2  -  x  -  i. 

(b)  x2  -  9x  -  10.  (j)  7x2  _  6x  -  1. 

(c)  1-X-X2.  (k)  3x2-5. 

(d)  4x2  -  4x  +  1.  (1)  2x2  +  X  -  8. 

(e)  5x2  +  10x+ 5.  (m)  2x2  +  x  +  8. 

(f)  3x2  -  5 X  -  22.  (n)  6x2  -  x  -  5. 

(g)  2  x2  +  13.  (o)  10  x2  +  60  X  +  90. 
(h)  9x2  -  6x  +  1.                                            (p)  7x^  +  7x  +  |. 

2.  For  what  real  values  of  the  parameter  k  will  one  root  of  each  of  the 
following  equations  have  the  value  assigned  ? 

One  root  to  he  zero : 

(a)  6x2+  5A;x-3Jfc2  4-3  =  0. 

(b)  2^-3x2  +  6x-fc2  +  3  =  0. 

(c)  x2  +  10x  + A:2  +  3  =  0. 

(d)  10x2  _  mx  +  3A;2  _  8A;  +  2  =  0. 
One  root  to  he  —  2: 

(e)  x2-2A;x  +  3  =  0. 

(f )  fcx2  -  X  +  3  fc2  -  1  =  0. 

(g)  fc2x2  +  6x  =  A:2-16. 
(h)  A;x2  +  2  fcx  =  -  3. 

(i)  10x2 -7/cx  +  A:2  + 9  =  0. 

3.  For  what  real  values  of  k  and  m  will  both  roots  of  each  of  the  following 
quadratic  equations  be  zero  ? 

(a)  5  x2  -|-  mx  -j-  A;  —  5  =  x. 

(b)  x2  +  (3  A;  -  m)  X  +  fc2  -  4  =  0. 

(c)  2x2  +  (7^2  +  l)x  +  A:2  =  0. 

(d)  x2  +  (m2  +  2fc-3m)x  +  4A;-6m  =  0. 

(e)  t^-{-{m'^  +  k'^-b)t  +  k  +  m  +  \  =  0. 

4.  For  what  real  values  of  the  parameter  are  the  roots  of  the  following 
equations  equal  ?    Verify  your  answers. 

(a)  A:x2  -  3x  -  1  =  0.  Ans.  A;  =  -  f . 

(b)  x2  -  A;x  +  9  =  0.  Am.  k  =  ±Q. 

(c)  2  A;x2  +  3  A:x  +  12  =  0.  Ans.  k  =  ^/. 

(d)  2  x2  +  A;x  -  1  =  0.  Ans.  None. 

(e)  5x2  -  3x  +  5  A:2  =  0.  ^^^  ^  ^  ^  _^_^^ 


Ans. 

k  =  ±l. 

Ans. 

A;  =  -  1  or  3. 

Ans. 

None. 

Ans. 

A:  =  f  ±  1  VTo. 

Ans. 

A;  =  -|. 

Ans. 

A:  =  - 1  or  -1 

Ans. 

None. 

Ans. 

None. 

Ans. 

k  =  -  7. 

Ans. 

A;  =  5,  m  =  1. 

Ans. 

k=±2,m  =  ± 

Ans. 

None. 

Ans. 

A;  =  0,  m  =  0. 

Ans. 

k  =  l,m  =  -2. 

k=-2,m  =  l. 

REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 


9 


Ans.   None. 

Ans.   A;  =  -  |. 

4  =  0. 

Ans.   6  =  —  4  or  1. 

:0. 

Ans.   m  =  —  1  or  2 

0. 

Ans.   None. 

Ans.   None. 

+  1=0. 

Ans.   None. 

0. 

Ans.    a  =  1  or  9. 

Ans.  p{p  —  2  km)  =  0. 

Ans.  p {m^p  —  2b)  =  0. 

Ans.  62  =  2  ahn. 

Ans.  b'^  =  r^{l  +  m2). 

Ans.  aWm"^  {k^ - a^m'^  +  62)  =  0. 

Ans.    b'^AB-\-m'^BC-\-AC=0. 


(f)  a;2  +  fcx  +  fc2  4.  2  =  0. 

(g)  x^-^2kx-k-\  =  0. 
(h)  x2  +  2  6x  +  2  62  +  3  6  - 

(i)  (m  +  2)x2-2ma;  +  l: 

(j)  (m2  +  4)x2  +  3x  +  2  = 
(k)  x2  +  (Z-3)x-l  =  0. 

(1)  (c2-8)y2_(2c-l)?/  + 
(ra)  a22  +  2  (a  +  3)  2;  +  16 

5.  Derive  the  equation  of  condition  in  order  that  the  roots  of  the  following 
equations  may  be  equal. 

(a)  m2x2  +  2fcmx-2px  =  -fc2. 

(b)  x2  +  2  mpx  +  2  6p  =  0. 

(c)  2mx2  +  2  6x  +  a2=:0. 

(d)  (1  +  m2)  x2  +  2  6mx  +  (62  -  r^)  =  0. 

(e)  (62  -  a2?n2)  y2  _  2  b'^ky  =  aWm"^  -  b%^. 

(f)  {A-^mW)x^-\-2bmBx-{-bW-^C  =  0. 

6.  For  what  real  values  of  the  parameter  do  the  common  solutions  of  the 
following  pairs  of  simultaneous  equations  become  identical  ? 

(a)  x  +  2y  =  fc,  x2  +  ?/2  =  5. 

(b)  y  =  mx  —  1,  x2  =  4  y. 

(c)  2x-?jy  =  b,  x^  +  2x  =  Sy. 

(d)  y  =  mx  +  10,  x^  +  y'^  =  10. 

(e)  Ix  +  y -2  =  0,  x^-Sy  =  0. 

(f)  X  +  4  y  =  c,  x2  +  2  2/2  =  9. 

(g)  x2  +  2/2  -  X  -  2  2/  =  0,  x  +  2y  =  c. 
(h)  x2  +  4  2/2  -  8  X  =  0,  mx  -  ?/  -  2  m  =  0. 

(i)  x2  +  2/2  -  fc  =  0,  3x  -  42/  =  25. 
(j)  x2  -  2/2  +  2  X  -  2/  =  3,  4  X  +  2/  =  c. 
(k)  2x2/  -  3x  -  2/  =  0,  y  +  Sx-\-k  =  0. 

(1)    x2  +  42/2-8  2/  =  0,    X  =  C. 
(m)   x2  +  42/2-82/  =  0,   2/  =  6. 

(n)  2x2  +  32/2  =  35,  4x  +  92/  =  fc. 

(0)  X2  +  X2/  +  2  X  +  2/  =  0,  2/  =  -  2  X  +  6. 

7.  If  the  common  solutions  of  the  following  pairs  of  simultaneous  equations 
are  to  become  identical,  what  is  the  corresponding  equation  of  condition  ? 

(a)  bx  +  ay  =  ab,  2/2  =  2px.  Ans.  ap  (2  62  +  ap)  =  0. 

(b)  2/  =  mx  +  6,  ^x2  +  52/  =  0.  Ans.  B (mW  -4bA)=0. 

(c)  y  =  m{x  -  a),  By^ -\- Dx  =  0.  Ans.  D (4  amW  -  D)  =  0. 

(d)  6x  +  ay  =  a6,  2x2/  +  c2  =  0.  ^ws.  a6(a6  +  2c2)  =  0. 

(e)  kx-y  =  c,  Ax^  +  By^  =  0.  ^ns.  c2^1?  -  kWC  -AC  =  0. 

(f)  X  cos  a  +  y  sin  a  =  p,  x2  +  2/2  =  r2.  ^ns.  p2  =  ^2, 


^ns. 

k=±5. 

-4  ns. 

m  =  ±l. 

J.ns. 

6  =  0. 

Ans. 

?M  =  ±  3. 

Ans. 

None. 

Ans. 

c  =  ±9. 

Ans. 

c  =  0  or  5. 

Ans. 

None. 

Ans. 

k=26. 

Ans. 

c  =  -  12  or  3. 

Ans. 

A:  =  -  6  or  0. 

An». 

c=±2. 

Ans. 

6  =  0,  2. 

Ans. 

A;  =  ±  35. 

Ans. 

6  =  —  4  or  0. 

10  ANALYTIC  GEOMETRY 

6.  Variables.  A  variable  is  a  quantity  to  which,  in  the  same 
investigation,  an  unlimited  number  of  values  can  be  assigned. 
In  a  particular  problem  the  variable  may,  in  general,  assume  any 
value  within  certain  limits  imposed  by  the  nature  of  the  problem. 
It  is  convenient  to  indicate  these  limits  by  inequalities. 

For  example,  if  the  variable  x  can  assume  any  value  between  —  2  and  5,  that 
is,  if  X  must  be  greater*  than  —  2  and  less  than  5,  the  simultaneous  inequalities 

x>-2,  a;<5, 
are  written  in  the  more  compact  form 

-2<a;<5. 

Similarly,  if  the  conditions  of  the  problem  limit  the  values  of  the  variable  x  to 
any  negative  number  less  than  or  equal  to  —  2,  and  to  any  positive  number  greater 
than  or  equal  to  5,  the  conditions 

a;<  —  2ora;  =  —  2,  and  a; > 5  or  x  =  5    ,, 
are  abbreviated  to  a;  <  —  2  and  a;  >  5. 

7.  Variation  in  sign  of  a  quadratic.  In  many  problems  it 
is  important  to  determine  the  algebraic  signs  of  the  results 
obtained  by  substituting  in  a  quadratic  different  values  for 
the  variable  unknown,  that  is,  to  determine  the  algebraic  signs 
of  the  values  of  a  quadratic  for  given  values  of  the  variable. 
The  discussion  of  this  question  depends  upon  the  definitions  of 
greater  and  less  already  given,  the  precise  point  necessary  being 
the  statement : 

If  a  is  a  given  real  constant  and  x  a  real  variable,  then 

J  when  x<a,  x  —  a  is  a  negative  number ; 
[when  x>a,  x  —  a  is  a  positive  number. 

By  the  aid  of  this  statement  and  the  identities  (7),  p.  4,  we 
easily  prove 

*  The  meaning  of  greater  and  less  for  real  numbers  (§  1)  is  defined  as  follows  :  a  is 
greater  than  b  when a-bisa,  positive  number,  and  a  is  less  than  b  when  a -fe  is  negative. 
Hence  any  negative  number  is  less  than  any  positive  number ;  and  if  a  and  b  are  both 
negative,  then  a  is  greater  than  b  when  the  numerical  value  of  a  is  less  than  the  numer- 
ical value  of  b. 

Thus  3<5,  but  -3>-5.  Therefore  changing  signs  throughout  an  inequality  reverses 
the  inequality  sign. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  11 

Theorem  III.  If  the  disGriminant  of  a  quadratic  is  positive,  the 
value  of  the  quadratic*  and  the  coefficient  of  the  second  power 
differ  in  sign  for  all  values  of  the  variable  lying  between  the  roots, 
and  agree  i7i  sign  for  all  other  values. 

If  the  discriminant  is  zero  or  negative,  the  value  of  the  quadratic 
and  the  coefficient  of  the  second  power  always  agree  in  sign. 

Proof  Denoting  the  variable  by  x,  and  writing  the  quadratic 
in  the  Typical  Form,  (1),  p.  2,  we  have,  by  (7),  p.  4, 

Case       I.   Ax^  -[-  Bx  +  C  =  A  (x  —  x-^  (x  —  x^  if  A  is  positive. 
Case     II.   Ax'^  -{-  Bx  -\-  C  =  A{x  —  x-^'^  if  A  is  zero. 

(a)  +  — 1  + ^  ^,  if  A 

i«  negciiive.  ^  ^ 

Consider  these  cases  in  turn. 

Case  I.  Since  the  roots  are  unequal,  let  x^  <  x^.  Then,  by 
(1),  we  have  at  once 

{x  —  iCi)  (x  —  X2)  is  negative  when  x^<.x<. x^, 
since  a:  —  Xi  is  positive,  and  x  —  x^\s>  negative ; 

{x  —  £Ci)  {x  —  CC2)  is  positive  when  xKx^oy  x>  x^, 
since  x  —  x^  and  x  —  x^  are  both  negative  or  both  positive. 

Therefore  the  quadratic  has  the  sign  of  —  ^  in  one  case,  and 
of  A  in  the  other. 

Case  II.  Since  {x  —  cci)^  is  positive  (p.  1),  the  sign  of  the 
quadratic  agrees  with  that  of  A. 

Case  III.  Since  A  is  negative,  ^  AC  —  B'^  =  —  ^  \^  positive  ; 
hence  the  expression  within  the  brackets  is  always  positive,  and 
the  sign  is  the  same  as  that  of  A.  q.e.i>. 

For  example,  consider  the  quadratic 

2  ^2  _  3  ^  +  1. 

Here  A  =  9-8=+l,  vl  =  2,  and  the  roots  are  \  and  1. 
'  ...  2r2-3<  +  l  =  2(i-  0(«-l). 

*It  is  assumed  that  all  the  numbers  involved  are  real.  Also,  since  the  value  of  the 
quadratic  is  zero  for  a  value  of  the  variable  equal  to  a  root,  any  such  value  of  the 
variable  is  excluded. 


12  ANALYTIC  GEOMETRY 

If  now  any  real  number  be  substituted  for  t  in  the  quadratic,  it  will  be 
found  that 

when  l<t<\,  the  quadratic  2 «2  _  3 ^  +  i< o ; 

when  i  <  i  or  i  >  1,  the  quadratic  2  <2  _  3  ^  4. 1  >  0. 

Again,  consider  the  quadratic  in  r, 

3  y2  +  4  r  +  9. 

Here  A  =  16  - 108  ==  -  92,  and  A  =  Z.  Hence,  by  Theorem  III,  if  any 
real  number  whatever  be  substituted  for  r,  the  result  will  always  be  a  posi- 
tive number. 

Applications  of  Theorem  III.  The  following  examples  illustrate  appli- 
cations of  Theorem  III. 

Ex.  1.  Determine  all  real  values  of  the  variable  for  which  the  following 
radicals  are  real. 


(a)  V3-  2x-x2;  (b)  V2 ^2  +  3 ?/  +  9. 

Solution.    Consider  the  quadratic  under  the  radical. 

In  (a),  A  =  4  +  12  =:  16,  ^  =  —  1,  and  the  roots  are  1  and  —  3. 

Applying  Theorem  III, 

when  —  3  < X <  1,  the  quadratic  3  —  2x  —  x2>0; 

when  x<  —  3orx>l,  the  quadratic  3  —  2 x  —  x^  < 0. 

Since  under  the  condition  of  the  problem  the  given  quadratic  must  be 
either  positive  or  zero,  we  have  —  3<x<l.     Ans. 

In  (b),  A  =  9  -  72  =  -  63,  and  A  =  2.  Hence,  by  Theorem  III,  the  quad- 
ratic is  positive,  and  therefore  the  square  root  is  real  for  every  real  value 
of  y.     Ans. 

Ex.  2,    For  what  values  of  the  parameter  k  are  the  roots  of  the  equation 

(2)  fcx2  +  2A:x-4x  =  2 -3^ 
(a)  real  and  unequal  ?     (b)  imaginary  ? 

Solution.    Writing  the  equation  in  the  Typical  Form, 

fcx2+  {2k-4)x  +  Sk-2  =  0, 
we  fiud 

(3)  A  =  B^  -  4  AC  =  -  S{k^  +  k  -  2). 

(See  Ex.  2,  p.  6.) 
By  Theorem  II,  p.  3, 

(a)  the  roots  are  real  and  unequal  if  -  8  (A;2  +  ^  -  2)  >  0 ; 

(b)  the  roots  are  imaginary  if  -  8  {k^  +  A;  -  2)  <  0. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  13 

Applying  Theorem  III  to  the  quadratic 

-S{k^  +  k-2), 
we  have,  since  A  =  64  +  512  =  576,  J.  =  —  8,  and  the  roots  are  —  2  and  1, 
when  -  2  <  A;  <  1,  the  quadratic  -  8  (A:^  +  fc  -  2)  >  0 ; 

when  A;  <  -  2  or  A;  >  1,  the  quadratic  -  8  (A;2  +  A;  -  2)  <  0. 
Hence 

(a)  the  roots  of  (2)  are  real  and  unequal  if  —  2  <  A;  <  1 ; 

(b)  the  roots  of  (2)  are  imaginary  ifA;<  —  2orA:>l.     -4ns. 

Ex.  3.    Show  that  the  simultaneous  equations 

(4)  y  =  mx  +  3 

(5)  ^  4x2  +  2/2  4-6x -16  =  0 

have  two  real  and  distinct  common  solutions  for  every  real  value  of  m. 

Solution.  Substituting  the  value  of  y  from  (4)  in  (5),  and  arranging  the 
result  in  the  Typical  Form,  we  get 

(6)  (4  +  m2)  x2  +  (6  771  +  6)  X  -  7  =  0. 

Calculating  the  discriminant  of  (6),  we  find,  neglecting  the  positive  factor  4, 

(7)  16  m2  +  18  m  +  37. 
Applying  Theorem  III,  p.  11,  to  the  quadratic  (7), 

A  =  324  -  64  •  37  is  negative,  A  =  16. 

Therefore  the  quadratic  (7)  has  a  positive  value  for  every  real  value  of  ?n, 
and  hence  the  roots  of  (6)  are,  by  Theorem  II,  p.  3,  always  real  and  unequal. 
That  is,  (6)  always  has  two  real  roots,  Xi  and  X2,  and  from  (4)  we  find  the 
corresponding  real  values  of  y,  namely,  i/i  and  2/2,  so  that  the  equations 
(4)  and  (5)  have  two  real  and  distinct  common  solutions,  (Xi,  yi),  (X2,  2/2), 
for  every  value  of  m.  q.e.d. 

PROBLEMS 

1.  Write  inequalities  to  express  that  the  values  of  the  variable  named  are 
limited  as  stated. 

(a)  X  has  any  value  from  0  to  5  inclusive. 

(b)  y  has  any  positive  value. 

(c)  t  has  any  negative  value. 

(d)  X  has  any  value  less  than  —  2  or  greater  than  —  1. 

(e)  r  has  any  value  from  —  3  to  8  inclusive. 

(f)  z  has  any  negative  value,  or  any  positive  value  not  less  than  3. 

(g)  X  has  any  value  not  less  than  —  8  nor  greater  than  2. 


14  ANALYTIC  GEOMETRY 

2.  Determine  the  sign  of  each  df  the  quadratics  of  the  first  problem  on 
p.  8  for  all  values  of  the  variable. 

3.  Determine  all  real  values  of  the  variable  for  which  the  square  root  of 
the  quadratics  of  problem  1,  p.  8,  are  real. 

4.  Determine  all  real  values  of  the  parameter  for  which  the  roots  of  each 
equation  of  problem  4,  p.  8,  are  (a)  real  and  unequal ;  (b)  imaginary. 

5.  In  problem  6,  p.  9,  find  all  real  values  of  the  parameter  in  each  case 
such  that  the  two  common  solutions  are  (a)  real  and  unequal ;  (b)  imaginary , 

6.  Determine  the  algebraic  sign  of  the  value  of  the  cubic 

2(x  +  l)(x-2)(x-4) 

for  any  value  of  the  variable. 

Hint.  In  this  case  the  roots  are  -1,  2,  4  in  the  order  of  magnitude.  Hence,  when 
x<—l,  each  factor  is  negative  [(1),  p.  10]  and  the  cubic  is  negative,  etc. 

Ans.  For  x  <  —  1,  cubic  <0;  — l<x<2,  cubic  >  0  j  2  <  x  <  4,  cubic  <  0  ; 
4  <  X,  cubic  >  0. 

7.  Determine  the  sign  of  the  value  of  each  of  the  following  quantics  for 
any  value  of  the  variable. 

Hint.  From  Algebra  we  know  that  any  quantic  with  real  coefficients  may  be 
resolved  into  real  factors  of  the  first  and  second  degrees.  The  sign  of  each  factor  for 
any  value  of  the  variable  may  then  be  determined  by  (1),  p.  10,  and  Theorem  III,  p.  11. 
It  is  well  first  to  arrange  the  real  roots  of  the  quantic  in  the  order  of  magnitude,  and 
then  it  is  necessary  to  consider  only  values  of  the  variable  less  than  any  root,  lying 
between  each  successive  pair,  and  greater  than  any  root,  as  in  problem  6. 

(a)  (X  +  1)  (2  x2-  4  X  +  7).  (f)  (x2  -  9)  (x2  -  16)  (x2  -  25). 

(b)  (x2-2x-3)(x3-4x2).  (g)  (3x2-12)  (2-x)(3-2x)(5x+4). 

(c)  (3x  +  8)(x2-4x  +  4)(x3-l).  (h)  (x  -  1)2(3  +  2x)(4  -  5x)  (6 -x)^. 

(d)  (2  x2  -f  3)  (x2  -  4)  (x4  -  1).  (i)  7  (x2  -  4)  (9  -  x2)  (16  -  x2). 

(e)  (2x  +  3)(x-l)(x  +  2)(x-3).  (j)  (x2  -  8)  (2x2  -  8)  (3x2  -  27). 

(k)  (2x  +  8)2(9-3x)(7-6x)(12-llx). 

8.  Infinite  roots.    Consider  the  quadratic  equation 

(1)  Ax"^  -{-Bx+C  =  0, 

whose  roots  are  x^  and  x^  [(3),  p.  3]. 
Then  the  equation 

(2)  Cx'^-{-Bx-{-A=0, 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  15 

obtained  from  (1)  by  reversing  tbe  order  of  the  coefficients,  has 

the  roots*  —  and  — j  that  is,  the  reciprocals  of  the  roots  of  (1). 

Let  us  now  fix  the  values  f  of  ^  and  C,  but  allow  A  to  dimin- 
ish indefinitely  in  numerical  value,  that  is,  allow  A  to  approach 

zero.     Then,  in  (2),  since  — .  (Theorem  I,  p.  3)  is  the  product  of 

the  roots,  this  product  must  also  approach  zero.  Therefore  one 
root  of  (2)  must  approach  zero ;  and  hence  its  reciprocal,  that  is, 
one  root  of  (1),  must  increase  indefinitely. 

Again,  let  us  in  (1)  and  (2)  fix  the  value  t  of  C  only,  and 
assume  that  both  B  and  A  approach  zero.     Then,  in  (2),  both  the 

B  A 

sum, )  and  the  product,  —  ?  of  the  roots  approach  zero,  and 

c  c 

hence  both  roots  also  approach  zero.  Hence  their  reciprocals, 
the  roots  of  (1),  must  increase  indefinitely. 

This  reasoning  establishes 

Theorem  IV.  If  the  coefficient  of  the  second  power  in  a  quadratic 
equation  is  variable  and  approaches  zero  as  a  limit,  then  one  root 
of  the  equation  becomes  infinite,  t  If  the  coefficient  of  the  first 
power  is  also  variable  and  approaches  zero  as  a  limit,  then  both 
roots  become  infinite. 

Ex.  1.  What  value  must  the  variable  k  approach  as  a  limit  in  order  that 
a  root  of  the  equation 

3x2  +  2A;x  -  A;2x2  -  3  -  2A:x2  =  0 
may  become  infinite  ? 

Solution.    Arranging  the  equation  in  the  Typical  Form,  we  have 

(A;2  +  2  fc  -  3)  x2  -  2  A:x  +  3  =  0. 
If  fc2  +  2  fc  —  3  =  0,  then  one  root  must  become  infinite.     Hence  k  must 
approach  1  or  —  3.     Ans. 

*  This  theorem  is  demonstrated  in  Algebra  and  may  be  easily  verified  thus  : 

The  equation  whose  roots  are  —  and  —  \s(x ^  (x ^  =  0. 

a?,  X2      \       Xi/  \       xj 

Multiplying  out  and  reducing,  this  becomes  x^x^  •  a;'  -  (aT^  +  a^gV  a:  +  1  =  0. 

O  R 

By  Theorem  I,  p.  3,  XyX^  =  - .  x^  +  x^  = ,  and  substitution  of  these  values  and 

multiplication  by  A  gives  (2). 

t  We  give  C  a  value  diif erent  from  zero. 

X  A  variable  whose  numerical  value  becomes  greater  than  any  assigned  number  is  said 
to  "  become  infinite." 


16  ANALYTIC  GEOMETRY 

Ex.  2.    What  values  must  k  and  m  approach  in  order  to  make  both  roots 
of  the  equation 

(62  _  a2?n2)  x2  _  2  a^kmx  -  a^k^  -  aW  =  0 
become  infinite  ? 

Solution.    By  Theorem  IV  we  must  have 

52  _  ^2^2  ^0,  or  fw  =  ±  - , 
and  2  a^km  =  0,  or    k  =  0. 

Hence  m  must  approach  +  -  or ,  and  k  must  approach  zero.     Ans. 


PROBLEMS 

1.  What  real  value  must  the  parameter  approach  as  a  limit  in  each  of  the 
following  equations  in  order  to  make  a  root  become  infinite  ? 

(a)  kz^-Sx-\-5  =  0.  (d)  (m^  -  i)x^  -  3x  +  8  =  0. 

(b)  (A:2  -  l)x2  +  6x  -  5  =  0.  (e)  (c2  -  3)y^  +  2cy  -  6  =  0. 

(c)  2  x2  -  3  X  +  A;2x2  +  5  =  kx^.  (f )  2  62^2  -  3  ?/  -  3  5?/2  +  2  =  -  2  2/2. 

2.  What  real  values  must  the  parameters  k  and  m  approach  in  order  that 
both  roots  of  each  of  the  following  equations  may  become  infinite  ? 

(a)  m2x2  ^(2k-m  +  l)x  +  6  =  0. 

(b)  (m2  -  3  m  +  2)  2/2  +  (3  fc  -  2  m)  y  +  2  =  0. 

(c)  (m2  ^  k'^  -  25)P  +  {m  -  7 k  +  2b)t  +  8  =  0. 

(d)  m2x2  4-  3A;x  +  A:2x2  -  4mx  +  25x  -  25x2  =  2. 

(e)  (m2  +  3)x2  +  (2 fc  -  5)x  +  8  =  0. 

8.  Equations  in  several  variables.  In  Analytic  Geometry  we 
are  concerned  chiefly  with  equations  in  two  or  more  variables. 

An  equation  is  said  to  be  satisfied  by  any  given  set  of  values 
of  the  variables  if  the  equation  reduces  to  a  numerical  equality 
when  these  values  are  substituted  for  the  variables. 


For  example,  x  =  2,y=—S  salasfy  the  equation 

«;2a;2  +  3?/2=35, 
since  2(2)2  +  3 ( -  3)2  =  35. 

Siitti\^rly,  x  =  —  1,  j/  =  0,-  z  =  —  4  satisfy  the  equation 
2a;2_3?y2_|.22_i8  =  o, 
since  2(- 1)2  _  3.O  +  (- 4)2  -  18  =  0. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  17 

An  equation  is  said  to  be  algebraic  in  any  number  of  variables, 
for  example  x,  y,  «,  if  it  can  be  transformed  into  an  equation 
each  of  whose  members  is  a  sum  of  terms  of  the  form  ax^^y^z^, 
where  a  is  a  constant  and  m,  n,p  are  positive  integers  or  zero. 

Thus  the  equations  x*  +  xhf'^  —  z^ -\-2x —  h  =  Q, 

x^y  +  2  x'hf^  =r_?/3  4.5a;2  +  2-a; 
are  algebraic. 

The  equation  x^  -\-y^  —  a^ 

is  algebraic. 

For,  squaring,  we  get    a;  +  2  x^y^  -{-y  —  a. 

Transposing,  2  xhj^  —  a  —  x  —  y. 

Squaring,  4iXy  =  a^  +  x^  -\-  y'^  —  2 ax  —  2 ay  ■}-  2xy. 

Transposing,  x^  -{-  y^  —  2xy  —  2 ax  —  2 ay  -{-  a^  ~  0.  q.k.d. 

The  degree  of  an  algebraic  equation  is  equal  to  the  highest 
degree  of  any  of  its  terms.''^  An  algebraic  equation  is  said  to 
be  arranged  with  respect  to  the  variables  when  all  its  terms  are 
transposed  to  the  left-hand  side  and  written  in  the  order  of 
descending  degrees. 

For  example,  to  arrange  the  equation 

2x'^  +  3y'-{-6x'—  2x'y'  —  2  +  x'^=  x'^y'—y'^ 
with  respect  to  the  variables  x',  y\  we  transpose  and  rewrite  the  terms  in  the  order 
x'^  —  x'Y  +  2aj'2 _  2xY  +  y'2  +  ex'  +  3?/'  —  2  =  0. 
This  equation  is  of  the  third  degree. 

An  equation  which  is  not  algebraic  is  said  to  be  transcendental. 

Examples  of  transcendental  equations  are 

y  =  sin  x,  y—  2^,  log  y  =  3  x. 

PROBLEMS 

1.  Show  that  each  of  the  following  equations  is  algebraic;  arrange  the 
terms  according  to  the  variables  x,  y,  or  x,  jsfeg,  and  determine  the  degree. 


(a)  x2  +  Vy-5  +  2  X  =  0. 

(b)  x^  +  y  +  3  X  =  0. 

(c)  xy  +  3  X*  +  6  x2?/  -  7  xy8  -f- 

(d)  X  +  y  +  z  4-  x^^;  -  3xy  -  2g2 

(e)  2/  =  2  +  Vx2  -  2  X  -  5. 

The  degree  of  any  term  is  the  sum  of  the  exponents  of  the  variables  in  that  term. 


18 


ANALYTIC  GEOMETRY 


(f)2/  = 

(g)  ^  = 

X  +  5  +  V2  x2  - 

-6x 

+  3. 
-Ey- 

-2/2. 

(h)  y  =  ^x  + -B  +  Vix2  +  ifx  +  iV". 

2.  Show  that  the  homogeneous  quadratic  * 
^x2  +  5X2/  +  C?/2 
may  be  written  in  one  of  the  three  forms  below  analogous  to  (7),  p.  4,  if 
the  discriminant  A  =  JB^  —  4  ^  (7  satisfies  the  condition  given  : 
Case     I.  Ax'^  +  Bxy  +  Gy'^  =  A{x-  hy)  (x  -  hy),  if  A  >  0 ; 
Case    II.  Ax^  +  Bxy  +  Cy^=A{x-  hyf,  if  A  =  0 ; 

r/  B     \2     4AC-B^ 


Case  III.  Ax^  +  Bxy  +  Cy^  =  A\  (x 


y)  + 


],. 


A<0. 


2A    /  4^2 

10.  Functions  of  an  angle  in  a  right  triangle.  In  any  right 
triangle  one  of  whose  acute  angles  is  A,  the  functions  of  A  are 
defined  as  follows  : 


sin  A  = 


opposite  side 


hypotenuse 

adiacent  side 

cos  A  =  -T^ — 7 : 

hypotenuse 

opposite  side 

adjacent  side 


tan^ 


csc^  = 


sec  A  = 


cot  A 


hypotenuse 
opposite  side 

hypotenuse 
adjacent  side 
adjacent  side 
opposite  side 


From  the  above  the  theorem  is  easily  derived : 

^        In  a  right  triangle  a  side   is  equal  to  the 
product  of  the  hypotenuse  and  the  sine  of  the 
angle  opposite  to  that  side,  or  of  the  hypote- 
a  nuse  and  the  cosine  of  the  angle  adjacent  to 
that  side. 


A         6        (7        11.  Angles    in    general 

an  angle  XOA  is  considered  as  gen- 
erated by  the  line  OA  rotating  from 
an  initial  position  OX.  The  angle  is 
positive  when  OA  rotates  from  OX 
counter-clockwise,  and  negative  when 
the  direction  of  rotation  of  OA  is 
clockwise. 


In    Trigonometry 


*  The  coefficients  A,  B,  C  and  the  numbers  l^,  Z,,  are  supposed  real. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  19^ 

The  fixed  line  OX  is  called  the  initial  line,  the  line  OA  the 
terminal  line. 

Measurement  of  angles.  There  are  two  important  methods 
of  measuring  angular  magnitude,  that  is,  there  are  two^unit 
angles. 

Degree  measure.  The  unit  angle  is  ^^^^  of  a  complete  revolu- 
tion, and  is  called  a  degree. 

Circular  measure.  The  unit  angle  is  an  angle  whose  subtend- 
ing arc  is  equal  to  the  radius  of  that  arc,  and  is  called  a  radian. 

The  fundamental  relation  between  the  unit  angles' is  given  by 
the  equation 

180  degrees  =  ir  radians  (tt  =  3.14159  •  •  •). 


Or  also,  by  solving  this, 

1  degree  =  j|^  =  .0174  . 

•  •  radians. 

180 
1  radian  =  ^^  =  57.29  • 

TT 

•  •  degrees. 

These  equations  enable  us  to  change  from  one  measurement  to 
another.  In  the  higher  mathematics  circular  measure  is  always 
used,  and  will  be  adopted  in  this  book. 

The  generating  line  is  conceived  of  as  rotating  around  0  through 
as  many  revolutions  as  we  choose.     Hence  the  important  result : 

Any  real  number  is  the  circular  measure  of  some  angle,  and 
conversely,  any  angle  is  measured  by  a  real  number. 

12.  Formulas  and  theorems  from  Trigonometry. 

Ill 

1.  cotx  = ;  secx  = ;  cscx  =  -; 

tan  X  cos  x  sin  x 

sin  a;        ,         cosx 

2.  tanx  = ;  cotx  = , 

cosx  sinx 

3.  sin2x  +  cos2x  =  1 ;  1  +  tan2x  =  sec^x ;  1  +  cot2x  =  csc2x. 

4.  sin  ( —  x)  =  —  sin  x ;  esc  ( —  x)  =  —  esc  x ; 
cos  ( —  x)  =  cos  X ;  sec  ( —  x)  =  sec  x ; 
tan (—  x)  =  —  tanx ;  cot (—  x)  =  —  cot x. 


20  ANALYTIC  GEOMETRY 

5.  sin  (tt  —  x)  =  sin x  ;  sin  {tt  -\-  x)  =  —  sin  x  ; 
cos  (tt  -  x)  =  —  cos  X  ;  cos  (tt  +  ic)  =:  —  cos  x  ; 
tan  (tt  —  x)  =  —  tan  x  ;  tan  (tt  +  x)  =     tan  x ; 

6.  sin  ( X  j  =  cos  X  ;  sin  (  --  +  x  j  ==  cos  x  ; 

cos  ( X  )  =  sin  X ;  cos  (  — j-  a^  )  =  —  sin  x  ; 

tan( X  )=  cotx;  tan(~- +  X  j  =  — cotx. 

7.  sin  (2  TT  -  x)  =  sin  (-  x)  =  -  sin  x,  etc. 

8.  sin  [x  -\-  y)  =  sin x  cos y  +  cos  x  sin  y. 

9.  sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y. 

10.  cos  (x  +  ?/)  =  cos X  cos  ?/  —  sin  x  sin  ?/.  '  ,  ; 

1 1 .  cos  (x  —  y)  =  cos  X  cos  2/  +  sin  x  sin  ^. 

tan  X  +  tan  y  10*/  \        tan  x  -  tan  y 

12.  tan  {x  +  y)= --^-        13.  tan  {x  -  y) - 


1  -  tan  X  tan  2/  1  +  tan  x  tan  y 

2tanx 


14.  sin  2  X  =  2  sin  x  cos  x  ;  cos  2  x  =  cos^  x  —  sin^  x  ;  tan  2  x 


1  -  tan2x 


X  ll  —  cosx  X  /l  +  cosx     ,      X  /I  — 

16.  sin-=±i^-^— ;cos-=±^-^— ;tan-=±^^ 


2\22  \22  \l  +  cosx 

16.    Theorem.    Law  of  sines.    In  any  triangle  the  sides  are  proportional 
to  the  sines  of  the  opposite  angles ; 


that  is, 


sin  A      sin  B      sin  C 


17.  Theorem.  Law  of  cosines.  In  any  triangle  the  square  of  a  side 
equals  the  sum  of  the  squares  of  the  two  other  sides  diminished  by  twice  the 
product  of  those  sides  by  the  cosine  of  their  included  angle ; 

that  is,  a2  _  52  _f.  c2  _  2  6c  cos^. 

18.  Theorem.  Area  of  a  triangle.  The  area  of  any  triangle  equals  one 
half  the  product  of  two  sides  by  the  sine  of  their  included  angle ; 

that  is,  area  =  ^  a6  sin  C  =  i  6c  sin  -4  =  ^  ca  sin  B. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  21 

13.  Natural  values  of  trigonometric  functions. 


Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

.0000 

0° 

.0000 

1.0000 

.0000 

00 

90° 

1.5708 

.0873 

5° 

.0872 

.9962 

.0875 

11.430 

85° 

1.4835 

.1745 

10° 

.1736 

.9848 

.1763 

5.671 

80° 

1.3963 

.2618 

15° 

.2588 

.9659 

.2679 

3.732 

75° 

1.3090 

.3491 

20° 

.3420 

.9397 

.3640 

2.747 

70° 

1.2217 

.4.363 

25° 

.4226 

.9063 

.4663 

2.145 

65° 

1.1345 

.5236 

30° 

.5000 

.8660 

.5774 

1.732 

60° 

1.0472 

.6109 

35° 

.5736 

.8192 

.7002 

1.428 

55° 

.9599 

.6981 

40° 

.6428 

.7660 

.8391 

1.192 

50° 

.8727 

.7854 

45° 

.7071 

.7071  . 

1.0000 

1.000 

45° 

.7854 

Cos 

Sin 

Cot 

Tan 

Angle  in 
Degrees 

Angle  in 
Radians 

Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

0 

0° 

0 

1 

0 

00 

1 

00 

It 

2 

90° 

1 

0 

GO 

0 

00 

1 

It 

180° 

0    • 

-1 

0 

00 

-1 

00 

2 

270° 

-1 

0 

CO 

° 

00 

-1 

2Tt 

360° 

0 

1 

0                       00 

1 

00 

22 


ANALYTIC  GEOMETRY 


Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

0 

0° 

0 

1 

0 

CO 

1 

CO 

Tt 

6 

30° 

1 

2 

V3 
2 

V3 
3 

V3 

2V3 
3 

2 

It 
4 

45° 

2 

2 

1 

1 

V2 

V2 

It 
3 

60° 

V3 

2 

1 
2 

V3 

V3 
3 

2 

2V3 
3 

Tt 

2 

90° 

1 

0 

CO 

0 

CO 

1 

14.  Rules  for  signs.. 

Quadrant 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

First    . "  .     .     . 

+ 

+ 

+ 

+ 

+ 

+ 

Second    .     .     . 

+ 

- 

- 

- 

- 

+ 

Third.     .     .     . 

- 

- 

+ 

+ 

- 

- 

Fourth     .     .     . 

- 

+ 

- 

- 

+ 

- 

9 


15. 

Greek  alphabet. 

Letters 

Names 

Letters 

Names 

A  a 

Alpha 

I      L 

Iota    / 

B^ 

Beta 

K    K 

Kappa 

r  7 

Gamma 

A  X 

Lambda 

A  5 

Delta 

M/i 

Mil 

E  e 

Epsilon 

N  V 

Nu 

z  f 

Zeta 

S  1 

Xi 

H^ 

Eta 

0  0 

Omicron 

e  d 

Theta 

n  IT 

Pi 

\r 

JU-t ^ 

^? 

Mf^  ■ 

Letters 

Names 

P^ 

Rho 

S    (T  S 

Sigma 

T    T 

Tau 

T  V 

Upsilon 

4.  0 

Phi 

Xx 

Chi 

^  1// 

Psi 

ft     U) 

Omega 

CHAPTER   II 
CARTESIAN    COORDINATES 

16,  Directed  line.  Let  X'X  be  an  indefinite  straight  line,  and 
let  a  point  0,  which  we  shall  call  the  origin  be  chosen  upon 
it.  Let  a  unit  of  length  be  adopted  and  assume  that  lengths 
measured  from  0  to  the  right  are  positive,  and  to  the  left  negative. 


-5-4-3-2-1    O-hl-t-2 +3  +  4+5  unit 


1 


Then  any  real  number  (p.  1),  if  taken  as  the  measure  of  the  length 
of  a  line  OP,  will  determine  a  point  P  on  the  line.  Conversely, 
to  each  point  P  on  the  line  will  correspond  a  real  number,  namely, 
the  measure  of  the  length  OP,  with  a  positive  or  negative  sign 
according  as  P  is  to  the  right  or  left  of  the  origin. 

The  direction  established  upon  X^X  by  passing  from  the  origin 
to  the  points  corresponding  to  the  positive  numbers  is  called  the 
positive  direction  on  the  line.    A  directed  line  is  a  straight  line  upon 


which  an  origin,  a  unit  of  length,  and  a  positive  direction  have 
been  assumed. 

An  arrowhead  is  usually  placed  upon  a  directed  line  to  indicate 
the  positive  direction. 

If  A  and  B  are  any  two  points  of  a  directed  line  such  that 

'     0A  =  a,   OB  =  b, 

then  the  length  of  the  segment  AB  is  always  given  hj  h  —  a;  that 
is,  the  length  oi  AB  is  the  difference  of  the  numbers  correspond- 
ing to  B  and  A.  This  statement  is  evidently  equivalent  to  the 
following  definition : 

23^ 


24  ANALYTIC  GEOMETRY 

For  all  positions  of  two  points  A  and  B  on  a  directed  line,  the 
length  AB  is  given  by 
(1)  AB  =  OB  -  OA, 

where  0  is  the  origin. 

(1)  (II)  (111)  (\Y) 

0   +3  -f-6       -4  0+3        -3       0        -h5       -6      -2    0 

6     A    B        B       0      A        A     0         B        B      A  0 

Illustrations. 

In  Fig.  I.  ^5  =05 -0^  =  6-3  =  4-3;^^  ==0^-0^=^3-6  =  - 3; 

II.  AB:=OB-OA^-^-?>  =  -l;  BA=  0^  -  Oi?  =  3- (- 4)  =+7; 

III.  ^i5=  OjB-0^  =  +  5-(-3)=  +  8;  j5.4=0^ -05  =- 3-5  =  -8; 

IV.  AB=  OJ5-0^=-6-(-2)  =  -4;  5^=0^-05  =  -2-(-6)= +4. 

The  following  properties  of  lengths  on  a  directed  line  are 
obvious :  ^^ 

(2)  AB  =  -BA. 

(3)  AB  is  positive  if  the  direction  from  A  to  B  agrees  with 
the  positive  direction  on  the  line,  and  negative  if  in  the  contrary 
direction. 

The  phrase  "  distance  between  two  points  "  should  not  be  used  if  these  points 
lie  upon  a  directed  line.  Instead,  we  speak  of  the  length  AB,  remembering  that 
the  lengths  AB  and  BA  are  not  equal,  but  that  AB  =  —  BA. 

17.  Cartesian*  coordinates.   Let  X'X  and  Y'Y  be  two  directed 
Yf  lines  intersecting  at  0,  and 

P  let  P  be  any  point  in  their 
plane.  Draw  lines  through 
P  parallel  to  X'X  and  TY 
respectively.     Then,  if 

OM  =  a,  0N  =  b, 
the  numbers  a,  b  are  called 
the  Cartesian  coordinates  of 
P,  a  the  abscissa  and  b  the 
ordinate.  The  directed  lines 
X'X  and  Y'Y  are  called  the 

*  So  called  after  Ren^  Descartes,  1596-1650,  who  first  introduqed  the  idea  of  coordinates 
into  the  study  of  Geometry. 


CARTESIAN  COORDINATES 


25 


axes  of  coordinates,  X^X  the  axis  of  abscissas,  Y^Y  the  axis  of 
ordinates,  and  their  intersection  O  the  origin. 

The  coordinates  a,  b  of  P  are  written  (a,  h),  and  the  symbol 
P  (a,  b)  is  to  be  read :  "  The  point  P,  whose  coordinates  are  a 
and  b:' 

Any  point  P  in  the  plane  determines  two  numbers,  the  coordi- 
nates of  P.  Conversely,  given  two  real  numbers  a'  and  b\  then 
a  point  P'  in  the  plane  may  always  be  constructed  whose  coordi- 
nates are  (a',  b^).  For  lay  off  OM'  =  a',  ON'  =  b',  and  draw  lines 
parallel  to  the  axes  through  AI'  and  N'.  These  lines  intersect  at 
P'(a',b').     Hence 

Every  point  determines  a  pair  of  real  numbers,  and  conversely, 
a  pair  of  real  numbers  determ,ines  a  point. 

The  imaginary  numbers  of  Algebra  have  no  place  in  this  repre- 
sentation, and  for  this  readon  elementary  Analytic  Geometry  is 
concerned  only  with  the  real  numbers  of  Algebra. 

18.  Rectangular  coordinates.  A  rectangular  system  of  coordi- 
nates is  determined  when  the  axes  X'X  and  Y'Y  are  perpendicular 


(oyTJ 


H   G) 


X' 


(10, 


(-5 


\-4} 


4) 


to  each  other.     This  is  the  usual  case,  and  will  be  assumed  unless 
otherwise  stated. 


26  ANALYTIC  GEOMETRY 

The  work  of  plotting  points  in  a  rectangular  system  is  much 
simplified  by  the  use  of  coordinate  or  plotting  paper,  constructed 
by  ruling  oft"  the  plane  into  equal  squares,  the  sides  being  parallel 
to  the  axes. 

In  the  figure,  p.  25,  several  points  are  plotted,  the  unit  of  length 
being  assumed  equal  to  one  division  on  each  axis.  The  method  is 
simply  this : 

Count  off  from  0  along  Z'Z  a  number  of  divisions  equal  to  the 
given  abscissa,  and  then  from  the  point  so  determined  a  number 
of  divisions  equal  to  the  given  ordinate,  observing  the 

Rule  for  signs : 

Abscissas  are  positive  or  negative  according  as  they  are  laid  off 
to  the  right  or  left  of  the  origin,  Ordinates  are 
positive  or  negative  according  as  they  are  laid 

Second        First       off  abovc  or  bclow  the  axis  of  X. 

^~'  ^  '     ^       Rectangular  axes  divide  the  plane  into  four 


X'         0 

Third 


-^  portions  called  quadrants ;  these  are  numbered 
(^,_)       as  in  the  figure,  in  which  the  proper  signs  of 
the  coordinates  are  also  indicated. 


PROBLEMS 

1.  Plot  accurately  the  points  (3,  2),  (3,  -  2),  (-  4,  3),  (6,  0),  (-  5,  0), 
(0,  4). 

2.  Plot  accurately  the  points  (1,  6),  (3,  -  2),  (-  2,  0),  (4,  -  3),  (-  7,  -  4), 
(-  2,  4),  (0,  -  1),  ( V3,  V2),  (-  V5,  0). 

3.  What  are  the  coordinates  of  the  origin?  Ans.  (0,  0). 

4.  In  what  quadrants  do  the  following  points  lie  if  a  and  h  are  positive 
numbers:   (-a,  6)?    (-a,  -6)?   (6,  -a)?    (a,  &)? 

5.  To  what  quadrants  is  a  point  limited  if  its  abscissa  is  positive  ?   nega- 
tive ?  its  ordinate  is  positive  ?   negative  ? 

6.  Plot  the  triangle  whose  vertices  are  (2,  —  1),  (—  2,  5),  (-  8,  —  4). 

7.  Plot  the  triangle  whose  vertices  are  (-  2,  0),  (5  V3  -  2,  5),  (-  2,  10). 

8.  Plot  the  quadrilateral  whose  vertices  are   (0,   —  2),    (4,  2),  (0,  6), 
(-4,2). 


p^ 


CARTESIAN  COORDINATES  27 

9.  If  a  point  moves  parallel  to  the  axis  of  x,  which  of  its  coordinates 
remains  constant  ?  if  parallel  to  the  axis  of  y  ? 

10.  Can  a  point  move  if  its  abscissa  is  zero  ?  Where  ?  Can  it  move  if  its 
ordinate  is  zero?  Where?  Can  it  move  if  both  abscissa  and  ordinate  are 
zero?   Where  will  it  be? 

11.  Where  may  a  point  be  found  if  its  abscissa  is  2?  if  its  ordinate 


12.  Where  do  all  those  points  lie  whose  abscissas  and  ordinates  are  equal  ? 

13.  Two  sides  of  a  rectangle  of  lengths  a  and  h  coincide  with  the  axes  of 
X  and  y  respectively.  What  are  the  coordinates  of  the  vertices  of  the  rec- 
tangle if  it  lies  in  the  first  quadrant  ?  in  the  second  quadrant  ?  in  the  third 
quadrant  ?  in  the  fourth  quadrant  ? 

14.  Construct  the  quadrilateral  whose  vertices  are  (—  3,  6),  (—  3,  0),  (3,  0), 
(3,  6).     What  kind  of  a  quadrilateral  is  it  ? 

15.  Join  (3,  5)  and  (-3,  -  5);  also  (3,  -  5)  and  (-  3,  5).  What  are  the 
coordinates  of  the  point  of  intersection  of  the  two  lines  ? 

16.  Show  that  (x,  y)  and  (x,  —  y)  are  symmetrical  with  respect  to  X'X] 
(x,  y)  and  (—  x,  ?/)  with  respect  toY'Y;  and  (x,  y)  and  (—  x,  —y)  with  respect 
to  the  origin. 

17.  A  line  joining  two  points  is  bisected  at  the  origin.  If  the  coordinates 
of  one  end  are  (a,  —  6),  what  will  be  the  coordinates  of  the  other  end  ? 


18.  Consider  the  bisectors  of  the  angles  between  the  coordinate  axes. 
What  is  the  relation  between  the  abscissa  and  ordinate  of  any  point  of  the 
bisector  in  the  first  and  third  quadrants  ?   second  and  fourth  quadrants  ? 

19.  A  square  whose  side  is  2  a  has  its  center  at  the  origin.  What  will  be 
the  coordinates  of  its  vertices  if  the  sides  are  parallel  to  the  axes  ?  if  the  diago- 
nals coincide  with  the  axes  ? 

Ans.   (a,  a),  (a,  —  a),  (—  a,  —  a),  (—  a,  a); 

(a  V2,  0),  (-  a  V2,  0),  (0,  a  V2),  (0,  -  a  V2). 

20.  An  equilateral  triangle  whose  side  is  a  has  its  base  on  the  axis  of  x 
and  the  opposite  vertex  above  X'X.  What  are  the  vertices  of  the  triangle  if 
the  center  of  the  base  is  at  the  origin  ?  if  the  lower  left-hand  vertex  is  at  the 

^„..(|.0),(-|,0),(0,^); 
(0,0),(a,0),(?,2^). 


28 


ANALYTIC  GEOMETRY 


19.  Angles.  The  angle  between  two  intersecting  directed  lines 
is  defined  to  be  the  angle  made  by  their  positive 
directions.  In  the  figuies  the  angle  between 
the  directed  lines  is  the  angle  marked  6. 

If  the  directed  lines  are  parallel,  then  the 
angle  between  them  is  zero  or  it  according  as 
the  positive  directions  agree  or  do  not  agree. 
Evidently  the  angle  between  two  directed 
lines  may  have  any  value  from  0  to  tt  inclusive. 
Eeversing  the  direction  of  either  directed  line 
changes  0  to  the  supplement  it  —  6.  If  both  directions  are 
reversed,  the  angle  is  unchanged. 


^  =  0 


e. 


When  it  is  desired  to  assign  a  positive  direction  to  a  line 
intersecting  X^X,  we  shall  always  assume  the  upward  direction 
as  positive  (see  figures). 


X  X' 


\B 


X 


CS) 


Theorem  1.    If  a  and  f3  are  the  angles  between  a  line  directed 
upward  and  the  rectangular  axes  OX  and  OY,  then 

(I)  cos  ^  =  sin  a. 

Froof.    The  figures  are  typical  of  all  possible  cases. 


In  Fig.  1, 
and  hence 


^-? 


cos  ^ 


2 
cos 


ex  I  =  sin  a.    (by  6,  p.  20) 


CARTESIAN  COORDINATES 


29 


In  Fig.  2, 
and  hence 

111  Fig.  3, 


cosyS 


/         7r\        . 
=  cos  [  a  —  —  I  =  sm  a. 

^  ^       (by  4  and  6,  p.  19) 


cos  y8  =  1  =  sin  or. 


Q.E.D. 


The  positive  direction  of  a  line  parallel  to  X'X  will  be  assumed 
to  agree  with  the  positive  direction  of  X'X,  that  is,  to  the  right. 

TT 

Hence  for  such  a  line  a  =  0,  p  =  —,  and  the  relation  (I)  still 
holds,  since 

TT 

cos  y8  =  cos  —  =  0  =  sin  0  =  sm  a. 


PROBLEMS 

1 .  Show  that  for  lines  directed  downward  cos  )3  =  —  sin  a. 

2.  What  are  the  values  of  a  and  jS  for  a  line  directed  N.E.  ?  N.  W.  ?  S.E.  ? 
S.W.  ?  (The  axes  are  assumed  to  indicate  the  four  cardinal  points  of  the 
compass.) 

3.  Find  the  relation  between  the  a's  and  /3's  of  two  perpendicular  lines 
directed  upward.  Ans.   a' 


20.  Orthogonal  projection.    The  orthogonal  projection  of  a  point 
upon  a  line  is  the  foot  of  the  perpendicular 


N 


let  fall  from  the  point  upon  the  line. 

Thus  in  the  figure 
M  is  the  orthogonal  projection  of  P  on  Z'X;         jj/ 
N  is  the  orthogonal  projection  of  P  on  FT;  X'    \       0  M  X 

P'  is  the  orthogonal  projection  of  P'on  X'X.         J 4iV 

If  A  and  B  are  two  points  of  a  directed 
line,  and  M  and  N  their  projections  upon  a 
second  directed  line  CD,  then  'MN  is  callSd  the  projection  of  AB 
upon  02>. 


r' 


30  ANALYTIC  GEOMETRY 

Theorem  II.    First  theorem  of  projection.    If  A  and  B  are  points 
upon  a  directed  line  making  an  angle  y  loith  a  second  directed  line 
CD,  then  the 
(II)         projection  of  the  length  AB  upon  CJD  =  AB  cos  y. 

•     Froof.    In  the  figures  let 

a  =  the  numerical  length  of  AB, 
I  =  the  numerical  length  of  ^^S"  or  BT; 
then  a  and  I  sue  positive  numbers  giving  the  lengths  of  the  respec- 
tive lines,  as  in  Plane  Geometry.     Now  apply  the  definition  of  the 
cosine  to  the  right  triangles  ABS  and  ABT  (p.  18). 

r.      A  ATT       ^   ^     ^     ^  ^     ^ 

B  B  T  II       N    M       N 

(1)  (2)  (3)  (4)  (5)  (6) 

In  Fig.  1,  1  =  a  cos  BA  S  =  a  cos  y, 

MN=l,  AB  =  a. 

.',  MN  =  AB  COS  y. 

In  Fig.  2,  I  =  a  cos  ABT  =  a  cos  (tt  —  y) 

=  —  a  cos  y,  (by  5,  p.  20) 

MN=^  I,  AB=—a.. 

.'.  MN  =  AB  cos  y. 

In  Fig.  3,  I  =  a  cos  ^  i^T"  =  a  cos  (tt  —  y) 

=  —  a  cos  y, 

MN  =  -l,  AB  =  a. 

.'.  MN  =  AB  cos  y.  ^ 

In  Fig.  4,  Z  =  a  cos  yl^r  =  «^  cos  y, 

MN  =  -l,  AB=-a. 

.-.  MN  =  AB  cos  y. 

In  Fig.  5,  y  =  0,  ikfiV  =  1,  AB  =  a. 

Hence  AfiV  =  ^jB  =  yl^  cos  0  (since  cos  0  =  1). 

.'.  MN  =  AB  cos  y. 

In  Fig.  6,  y  =  TT,  iV/i\r  =  -l,  AB=a. 

Hence  MN^=  —  AB  =  AB  cos  tt  (since  cos  tt  =  —  1). 

.'.  ikfiV  =  ^5c0Sy.  Q.E.D. 


CARTESIAlST  COORDINATES 


31 


Consider  any  two  given  points 

Pi{^i,  2/1) »  -^2(^2,  2/2)- 

Then  in  the  figure 

7I/1M2  =  projection  of  PxP^  on  X'X, 
NiNi  =  projection  of  PiPg  on  Y'Y. 


Pi(xi,y,) 


■^  fa:  2, 2/2)  y, 

Butby  (1),  p.  24, 

M1M2  =  OM^  -  OMi  =  0^2  -  «i, 
N^N^  =  ON2  -  ON^  =  ^2  -  2/1- 
Hence 
Theorem  III.    Given  amj  tivo points  P^ix^,  y^,  ^2(2^2,  2/2);  ^^^^ 

a?2  —  a?!  =  projection  of  P^P^  on  X'X; 
projection  of  PxP^  on  F'  F. 


(Ill) 


21.  Lengths.    We  may  now  easily  prove  the  important 
Theorem  IV.    The  length  I  of  the  line  Y 

joining  two  points  P^  (a?i,  y^,  P^  {x^,  2/2) 
is  given  by  the  formula  jy- 

(IV)  I  =  V(xi-X2)^+(2/i-2/2)'- 

Proof    Draw  lines  through  Pi  and  -^ 
P2  parallel  to  the  axes  to  form  the 
right  triangle  PiSP^.  Y 

Then  SP^  =  M^M^  =  x^  -  x^, 

P2S  =  N^N^  =  2/1  -  2/2> 
PiP2=V^^'4-^'; 
arid  hence 


?AC«2,2/2) 


Z  =  V(a;i-(r2)'+(2/i-2/2)'. 


(by  III) 
(by  III) 

Q.E.D. 


32 


ANALYTIC  GEOMETEY 


The  method  used  in  deriving  (IV)  for  any  positions  of  Pi  and 
P2  is  the  following  : 

Construct  a  right  triangle  by  drawing  lines  parallel  to  the  axes 
through  Pi  and  P^-  'J-'he  sides  of  this  triangle  are  equal  to  the 
projections  of  the  length  P1P2  upon  the  axes.  But  these  projec- 
tions are  always  given  by  (HI)?  or  by  (III)  with  one  or  both 
signs  changed.  The  required  length  is  then  the  square  root  of 
the  sum  of  the  squares  of  these  projections,  so  that  the  change  in 
sign  mentioned  may  be  neglected.  A  number  of  different  figures 
should  be  drawn  to  make  the  method  clear. 

Ex.  1.    Find  the  length  of  the  hne  joinmg  the  points  (1,  3)  and  (—5,  5). 

Solution.    Call  (1,  3)  Pi,  and  (-  5,  5)P2. 
Then 

^1  =  1»  2/1  =  3,  and  Xa  =  —  5,  2/2  =  5; 
and  substituting  in  (IV),  we  have 

Vio 


Y 

(-5 

I^ 

<, 

--> 

^ 

""^ 

>. 

"-., 

(1, 

3) 

0 

Y 

1  =  V{1  +  5)^  +  (3  -  5)^ 


2  VlO. 


X 


It  should  be  noticed  that  we  are  simply- 
finding  the  hypotenuse  of  a  right  triangle 
whose  sides  are  6  and  2. 


Remark.  The  fact  that  formulas  (III)  and  (lY)  are  true  for  all 
positions  of  the  points  Pi  and  P2  is  of  fundamental  importance. 
The  application  of  these  formulas  to  any  given  problem  is  there- 
fore simply  a  matter  of  direct  substitution,  as  the  example  worked 
out  above  illustrates.  In  deriving  such  general  formulas,  since 
it  is  immaterial  in  what  quadrants  the  assumed  points  lie,  it  is 
most  convenient  to  draw  the  figure  so  that  the  points  lie  in  the 
first  quadrant,  or,  in  general,  so  that  all  the  quantities  assumed 
as  known  shall  he  positive. 


PROBLEMS 

1.  Find  the  projections  on  the  axes  and  the  length  of  the  lines  joining 
the  following  points : 

(a)  (-  4,  —  4)_and  (1,  3).      _       Ans.  Projections  5,  7;  length  =  Vri. 

(b)  (- V2,  V3)  and  (V3,  V2). 

Ans.  Projections  V3  +  V2,  V2  -  V3,-  length  =  VlO. 


CARTESIAN  COORDINATES  33 

(c)  (0,  0)  and  (  - ,  -— —  j .  Ans.  Projections  - ,  -  ^3 ;  length  =  a. 

(d)  {a -\- h^  c  ■\-  a)  and  (c  +  a,  6  +  c). 


" 


Ans.    Projections  c  —  h^h  —  a\  length  =  ■>/(&  —  c)2  +  (a  —  &)2, 

2.  Find  the  projections  of  the  sides  of  the  following  triangles  upon  the 

axes: 

(a)  (0,  6),  (1,  2),  (3,  -  5). 

(b)  (1,  0).  (-1,  -6),  (-1,  -8). 

(c)  (a,  6),  (6,  c),  (c,  d). 

3.  Find  the  lengths  of  the  sides  of  the  triangles  in  problem  2. 

4.  Work  out  formulas  (III)  and  (IV),  (a)  if  Xi  =  Xg;  (b)  if  y^  —  yi. 

5.  Find  the  lengths  of  the  side;S'of  the  triangle  whose  vertices  are  (4,  3), 
(2, -2),  (-3,  5).  /  • 

6.  Show  that  the  points  (1,  4),  (4, 1),  (5,  5)  are  the  vertices  of  an  isosceles 
triangle. 

7.  Show  that  the  points  (2,  2^,  (-2,  -  2),  (2  Vs,  -  2  V3)  are  the  vertices 
of  an  equilateral  triangle.         / 

8.  Show  that  (3,  0),  (6,  4),  (—  1,  3)  are  the  vertices  of  a  right  triangle. 
What  is  its  area  ? 

9.  Prove  that  (-  4,  -  2),  (2,  0),  (8,  6),  (2,  4)  are  the  vertices  of  a  paral- 
IMogram.     Also  find  the  lengths  of  the  diagonals. 

10.  Show  that  (11,  2),  (6,  -  10),  (-6,  -  5),  (-  1,  7)  are  the  vertices  of 
a  square.     Find  its  area. 

11.  Show  that  the  points  (1,  3),  (2,  Vo),  (2,  —  V6)  are  equidistant  from 
the  origin,  that  is,  show  that  they  lie  on  a  circle  with  its  center  at  the  origin 
and  its  radius  VlO. 

12.  Show  that  the  diagonals  of  any  rectangle  are  equal. 

13.  Find  the  perimeter  of  the  triangle  whose  vertices  are  (a,  6),  (—  a,  6), 
(-a,  -6). 

14.  Find  the  perimeter  of  the  polygon  formed  by  joining  the  following 
points  two  by  two  in  order : 

(6,  4),  (4,  -  3),  (0,  -  1),  (-  5,  -  4),  (-  2,  1). 

16.  One  end  of  a  line  whose  length  is  13  is  the  point  (-4,  8);  the  ordi- 
nate of  the  other  end  is  3.     What  is  its  abscissa  ?  Ans.  8  or  —  16. 

16.  What  equation  must  the  coordinates  of  the  point  (x,  y)  satisfy  if  its 
distance  from  the  point  (7,  —  2)  is  equal  to  11  ?  •        . 


34  ANALYTIC  GEOMETRY 

17.  What  equation  expresses  algebraically  the  fact  that  the  point  (x,  y)  is 
equidistant  from  the  points  (2,  3)  and  (4,  5)? 

18.  If  the  angle  XOY  (Fig.,  p.  24)  equals  w,  show  that  the  length  of  the 
line  joining  Pi(xi,  ?/i)  and  P^ix^,  y-i)  is  given  by 


I  =  V(xi  -  Xa)^  +  (?/i  -  2/2)^  +  2  (Xi  -  X2)  (z/i  -  2/2)  cos  w. 

19.  If  w  =  — ,  find  distance  between  the  points  (—3,  3)  and  (4,  —  2). 

Ain^.   V39. 

20.  If  w  =  — ,  find  the  perimeter  of  the  triangle  whose  vertices  are  (1,  3), 
(2,  7),  (-  4,  -  4).  An8,   V2I  +  V223  +  Vl09. 

21.  If  w  =  -,  find  the  perimeter  of  triangle  (1,  2),  (-  2,  -  4),  (3,  -  5). 

Aus.  3  V5  +  2  V3  +  V26  -  5  V3  +  V53  -  14  V3. 

22.  Prove  that  (6,  6),  (7,  -  1),  (0,  -2),  (-2,  2)  lie  on  a  circle  whose 
center  is  at  (3,  2). 

23.  If  w  =  ^ — ,  find  the  distance  between  ( Vs,  V2),  (-  Vi,  V3). 

Ans.    VlO  +  V2. 

24.  Show  that  the  sum  of  the  projections  of  the  sides  of  a  polygon  upon 
either  axis  is  zero  if  each  side  is  given  a  direction  established  by  passing 
continuously  around  the  perimeter. 

22.  Inclination  and  slope.  The  inclination  of  a  line  is  the  angle 
between  the  axis  of  x  and  the  line  when  the  latter  is  given  the 

upward  direction  (p.  28). 

The  slope  of  a  line  is  the  tangent  of 
its  inclination. 

The  inclination  of   a  line  will  be 
J         denoted  by  a^  ai,  org?  «'?  etc. ;  its  slope 
-^ — :>  by  m,  m^  mg,  m',  etc.,  so  that  m  =  tan  a, 
mi  =  tan  ai,  etc. 

The  inclination  may  be  any  angle 
from  0  to  TT  inclusive  (p.  28).      The 


T/N 


r' 


slope  may  be  any  real  number,  since  the  tangent  of  an  angle  in 
the  first  two  quadrants  may  be  any  number  positive  or  negative. 
The  slope  of  a  line  parallel  to  X'X  is  of  course  zero,  since  the. 
inclination  is  0  or  tt.    For  a  line  parallel  to  Y'  Y  the  slope  is  infinite. 


CARTESIAN  COORDINATES 


35 


Theorem  V.    The  slope  tn  of  the  line  passing  through  two  points 
Pii^ij  yi)}  Pii^i,  y^  is  given  by 

Uy-t    Uyo 


Proof. 


(1) 


Similarly, 


(2) 


But 


M1M2  =  0*2  —  iCj 

=  P1P2  COS  a. 
.'.  P1P2  cos  a  =  X2  —  Xi. 

^1^2  =  2/2-2/1 

=  P1P2  cos  /3. 

.'.  P^P^  cos  p  =  1/2- yi- 
cos  (3  =  sin  a. 


(by  (III),  p.  31) 
(by  (II),  p.  30) 


(by  (III),  p.  31) 
(by  (II),  p.  30) 

(by  (I),  p.  28) 


Hence,  from  (2), 


(3) 


Dividing  (3)  by  (1),     tan  a  =  m  = 


PyP^.  sin  a  =  yz  —  y^. 

^2   -  2/1  _ 

x^  —  x^ 

Remark.  Formula  (V)  may  be  verified  by 
constructing  a  right  triangle  whose  hypot- 
enuse is  P1P2)  s-s  on  p.  31,  whence  tan  a 
(=  tan  Z  SP1P2)  is  found  directly  as  the  ratio 
of  the  opposite  side,  SP^  =  2/2  —  2/i)  ^^  t^® 
adjacent  side,  PiS  =  X2  —  Xi* 


2/1-2/2 


Q.E.D. 


Y' 

P2/ 

y---^s 

Am. 

0 

/                ^ 

*  To  construct  a  line  passing  through  a  given  point  Pi  whose  slope  is  a  positive  frac- 

Pj  a  units  above  S,  and 
must  lie  to  the  left  of  P^ 


tiori  - ,  we  mark  a  point  S  h  units  to  the  right  of  P^  and  a  point 
draw  PiPg-  If  the  slope  is  a  negative  fraction,  —  ^ ,  then  either  S 
or  Pj  niust  lie  below  S. 


86  ANALYTIC  GEOMETRY 

Theorem  VI.  If  two  lines  are  parallel,  their  slopes  are  equal;  if 
perpendicular,  the  slope  of  one  is  the  negative  reciprocal  of  the  slope 
of  the  other,  and  conversely. 

Proof  Let  a-^  and  a^  be  the  inclinations  and  m^  and  m^^  the 
slopes  of  the  lines. 

If  the  lines  are  parallel,  a^  =  a^.     .'.  m^  =  mg. 

If  the  lines  are  perpendicular,  as  in  the  figure, 

IT  TT 

«'2  =  ^1  +  2'   °^  a^  =  oc^-  ^^ 


tan  ( cx^  —  -^  j 

(by  4  and  6,  p.  19) 
(by  1,  p.  19) 


1 

Q.E.D. 
2 


m 

The  converse  is  proved  by  retracing  the  steps  with  the  assump 
tion,  in  the  second  part,  that  ^2  is  greater  than  aj. 

PROBLEMS 

1.  Find  the  slope  of  the  line  joining  (1,  3)  and  (2,  7).  Ans.   4. 

2.  Find  the  slope  of  the  line  joining  (2,  7)  and  (—4,  —  4).        Ans.   y-. 

3.  Find  the  slope  of  the  line  joining  (Vs,  V2)  and  (—  V2,  Vs). 

Ans.  2V6-5. 

4.  Find  the  slope  of  the  line  joining  {a  +  h,  c  -\-  a),  (c  -{-  a,  b  +  c). 

b  —  a 


Ans. 


c-b 


5.  Find  the  slopes  of  the  sides  of  the  triangle  whose  vertices  are  (1,  1), 
(-  1,  -  1),  ( V3,  -  V3).  ^^^    ^^  1+V3^  I-V3 

'  i-Vs'  1+V3 

6.  Prove  by  means  of  slopes  that  (-  4,  -  2),  (2,  0),  (8,  6),  (2,  4)  are  the 
vertices  of  a  parallelogram. 

7.  Prove  by  means  of  slopes  that  (3,  0),  (6,  4),  (—  1,  3)  are  the  vertices 
of  a  right  triangle. 

8.  Prove  by  means  of  slopes  that  (0,  -2),  (4,  2),  (0,  G),  (-4,  2)  are  the 
vertices  of  a  rectangle,  and  hence,  by  (IV),  of  a  square. 


UNIVERSITY 

OF 

^_    v^T>4_,«^j,gj^^  COORDINATES  37 

9.  Prove  by  means  of  their  slopes  that  the  diagonals  of  the  square  in 
problem  8  are  perpendicular. 

10.  Prove  by  means  of  slopes  that  (10,  0),  (5,  6),  (5,  -  5),  (-  5,  5)  are 
the  vertices  of  a  trapezoid. 

11.  Show  that  the  line  joining  (a,  b)  and  (c,  -  d)  is  parallel  to  the  line 
joining  (-a,  —  h)  and  (-  c,  d). 

12.  Show  that  the  line  joining  the  origin  to  (a,  b)  is  perpendicular  to  the 
line  joining  the  origin  to  (—6,  a). 

13.  What  is  the  inclination  of  a  line  parallel  to  Y'Y?    perpendicular  to 
Y'Y? 

14.  What  is  the- slope  of  a  line  parallel  to  Y'Y?    perpendicular  to  Y'Y? 

15.  What  is  the  inclination  of  the  line  joining  (2,  2)  and  (—2,  —  2)? 

Ans.  —' 
4 


16.  What  is  the  inclination  of  the  line  joining  (—  2,  0)  and  (—  5,  3)? 

Ans. 

17.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (4,  V3)  ? 

Am 

18.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (2,  Vs)  ? 


Ans.  —-' 
4 


A  ^ 

Ans.  - 


27t 

Ans.—- 


19.  What  is  the  inclination  of  the  line  joining  (0,  —  4)  and  (—  V3,  —  5)  ? 


Ans.  ^- 


20.  What  is  the  inclination  of  the  line  joining  (0,  0)  and  (—  V3,  1) 


57r 


Ans.    ^, 

21.  Prove  by  means  of  slopes  that  (2,  3),  (1,  —  3),  (3,  9)  lie  on  the  same 
straight  line. 

22.  Prove  that  the  points  (a,  6  +  c),  (6,  c  -\-  a),  and  (c,  a  +  6)  lie  on  the 
same  straight  line. 

23.  Prove  that  (1,  5)  is  on  the  line  joining  the  points  (0,  2)  and  (2,  8) 
and  is  equidistant  from  them. 

24.  Prove  that  the  line  joining  (3,  —  2)  and  (5,  1)  is  perpendicular  to  the 
line  joining  (10,  0)  and  (13,  -  2). 


38  ANALYTIC  GEOMETRY 

23.  Point  of  division.    Let  Pi  and  P^  be  two  fixed  points  on  a 
directed  line.     Any  third  point  on  the  line,  as  P  or  P\  is  said 


Pi  P         P2  P' 

"to  divide  the  line  into  two  segments,"  and  is  called  a  point 
of  division.  The  division  is  called  internal  or  external  according 
as  the  point  falls  within  or  without  PiP^-  The  position  of  the 
point  of  division  depends  upon  the  ratio  of  its  distances  from  Pj 
and  Pa-  Since,  however,  the  line  is  directed,  some  convention 
must  be  made  as  to  the  manner  of  reading  these  distances.  We 
therefore  adopt  the  rule  : 

If  P  is  a  point  of  division  on  a  directed  line  passing  through 
Pi  and  P2,  then  P  is  said  to  divide  P1P2  into  the  segments  PiP 

P  P 
and  PP2.     The  ratio  of  division  is  the  value  of  the  ratio*  -^— • 

We  shall  denote  this  ratio  by  X,  that  is,  ^ 

If  the  division  is  internal,  PiP  and  PP2  agree  in  direction  and 
therefore  in  sign,  and  X  is  therefore  positive.  In  external  divi- 
sion X  is  negative.  The  sign  of  A.  therefore  indicates  whether 
the  point  of  division  P  is  within  or  without  the  segment  P1P2 ; 
and  the  numerical  value  determines  whether  P  lies  nearer  Pi 
or  Pg.     The  distribution  of  X  is  indicated  in  the  figure. 

-1<\<0  XrO  X>0  X=00        -<X><X<-1 


Px  p. 

That  is,  X  may  have  any  positive  value  between  Pi  and  Pg,  any 
negative  value  between  0  and  —  1  to  the  left  of  Pi,  and  any  nega- 
tive value  between  —  1  and  —  00  to  the  right  of  P2.  The  value 
—  1  for  X  is  excluded. 

*  To  assist  the  memory  in  writing  down  this  ratio,  notice  that  the  point  of  division  P 
is  written  last  in  the  numerator  and  first  in  the  denominator. 


CARTESIAN  COORDINATES 


39 


Introducing  coordinates,  we  next  prove 

Theorem  VII.  Point  of  division.  The  coordinates  (x,  y)  of  the 
point  of  division  P  on  the  line  joining  P^  (x^,  y^,  P^  (x^,  y^),  such 
that  the  ratio  of  the  segments  is 


rp. 


=  A, 


are  given  hy  the  formulas 
(VII) 


iC  =   — : ; — > 


1  +  A 


y  = 


Vi  +  A2/2 


l  +  A 


Proof.    Given 


X  = 


PP. 


Let  a  be  the  inclination  of  the  line  PiP 
upon  the  axis  of  x. 

Then,  by  the  first  theorem  of  projection  [(II),  p.  30], 

MiM  =  P^P  cos  or, 
MM2  =  PP2  cos  a. 
M^M  _  P,P  _ 
MM2~  PP.~  ' 
M^M  =  X  —  x^j 


Mi    M  M.X 


Project  Pi,  P,  P2 


Dividing, 
But 


Substituting, 


MM. 


X. 


(by  hypothesis) 
(by  (III),  p.  31) 


A. 


Clearing  of  fractions  and  solving  for  x, 

x^  -f-  Aa^g 


Similarly, 


y 


Q.E.D. 


1  +  A 

yi  -h  A..?/2 
1  +  A   * 

Corollary.  Middle  point.  The  coordinates  (x,  y)  of  the  middle 
point  of  the  line  joining  Pi(xi,  y^),  P^ix^,  2/2)  are  found  hy  taking 
the  averages  of  the  given  abscissas  ^and  ordinates ;  that  is, 

-  |(a?i  +  ajg),  y  =  | (y^  +  y^). 


3C 


P,P 


For  if  P  is  the  middle  point  of  P^P^,  then  A  =  — ^  =  1. 


40 


ANALYTIC  GEOMETRY 


Ex.  1.  Find  the  point  P  dividing  Pi  (-1,  -6),  P^  (3,  0)  in  the  ratio  \=-\. 
Solution.  Applying  (VII),  xi  =  -  1,  yi=—  6, 
=  3,  y2  =  0. 


X  = 


y  = 


-1- 

i 

3 

1- 

-6- 

■0 

f 

-6_ 

T" 

Ans. 


Hence  Pis  (-  2i,  -8). 

Ex.  2.  Find  the  coordinates  of  the  point  of 
intersection  of  the  medians  of  a  triangle  whose 
vertices  are  (xi,  yi),  (X2,  ^2),  (a^s,  2/3). 

Solution.  By  Plane  Geometry  we  have  to  find 
the  point  P  on  the  median  AD  such  that  AP  =  |  AD,  that  is,  AP :  PD  : :  2  : 1, 
or  X  =  2. 

By  the  Corollary,  D  is  [|(X2  +  X3),  i (2/2  +  Vs)]- 
To  find  P,  apply  (VII),  remembering  that  A  corre- 
sponds to  (xi,  2/1)  and  D  to  (X2,  2/2)- 


AC^i^Vi) 


This  gives  x 


2/  = 


Xi  +  2  •  i  (X2  +  X3) 

2-i(2/2  +  ?/3) 


2/1 


.     r3J3,2/3>) 


1  +  2 
.-.  X  =  1  (xi  +  X2  +  X3),  2/  =  I  (2/1  +  2/2  +  2/3)-     ^ns. 

Hence  the  abscissa  of  the  intersection  of  the  medians  of  a  triangle  is  the 
average  of  the  abscissas  of  the  vertices,  and  similarly  for  the  ordinate. 

The  symmetry  of  these  answers  is  evidence  that  the  particular  median 
chosen  is  immaterial,  and  the  formulas  therefore  prove  the  fact  of  the  intersec- 
tion of  the  medians. 


PROBLEMS 

1.  Find  the  coordinates  of  the  middle  point  of  the  line  joining  (4,  —  6) 
and  (-2,  -4).  Ans.    (1,  -  5). 

2.  Find  the  coordinates  of  the  middle  point  of  the  line  joining  {a-\-b,c  +  d) 
and  (a  —  6,  d  —  c).  Ans.    (a,  d). 

3.  Find  the  middle  points  of  the  sides  of  the  triangle  whose  vertices  are 
(2,  3),  (4,  —  5),  and  (—3,  —  6) ;  also  find  the  lengths  of  the  medians. 

4.  Find  the  coordinates  of  the  point  which  divides  the  line  joining  (—1,4) 
and  (—5,  —  8)  in  the  ratio  1 :  3.  Ans.    (—  2,  1). 

6.  Find  the  coordinates  of  the  point  which   divides  the  line  joining 
(-3,  -5)  and  (6,  9)  in  the  ratio  2:5.  Ans.    (-  f,  -1). 


CARTESIAN  COORDINATES  41*^^ 

6.  Find  the  coordinates  of  the  point  which  divides  the  line  joining  (2,  6) 
and  (—  4,  8)  into  segments  whose  ratio  is  —  |.  Ans.    {—  22,  14). 

7.  Find  the  coordinates  of   the  point   which  divides  the  line  joining 
(—3,  —  4)  and  (5,  2)  into  segments  whose  ratio  is  —  f.   Ans.  (—19,  —  16). 

8.  Find  the  coordinates  of  the  points  which  trisect  the  line  joining  the 
points  (-  2,  -  1)  and  (3,  2).  Ans.    (-  ^,  0),  (4,  1). 

9.  Prove  that  the  middle  point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices. 

10.  Show  that  the  diagonals  of  the  parallelogram  whose  vertices  are  (1,2), 
(-  5,  -  3),  (7,  -  6),  (1,  -  11)  bisect  each  other. 

11.  Prove  that  the  diagonals  of  any  parallelogram  mutually  bisect  each 
other. 

12.  Show  that  the  lines  joining  the  middle  points  of  the  opposite  sides  of 
the  quadrilateral  whose  vertices  are  (6,  8),  (—  4,  0),  (—  2,  —  6),  (4,  —  4)  bisect 
each  other. 

13.  In  the  quadrilateral  of  problem  12  show  by  means  of  slopes  that  the  ■* 
lines  joining  the  middle  points  of  the  adjacent  sides  form  a  parallelogram. 

14.  Show  that  in  the  trapezoid  whose  vertices  are  (—  8,  0),  (—4,  —  4), 
(—4,  4),  and  (4,  —  4)  the  length  of  the  line  joining  the  middle  points  of  the 
non-parallel  sides  is  equal  to  one  half  the  sum  of  the  lengths  of  the  parallel 
sides.     Also  prove  that  it  is  parallel  to  the  parallel  sides. 

15.  In  what  ratio  does  the  point  (—2,  3)  divide  the  line  joining  the  points 
(-3,  5)  and  (4,  -9)?  Ans.    |. 

16.  In  what  ratio  does  the  point  (16,  3)  divide  the  line  joining  the  points 
(-5,  0)and  (2,  1)?  Ans.    -  |. 

17.  Given  the  triangle  whose  vertices  are  (—  5,  3),  (1,  —  3),  (7,  6);  show 
that  a  line  joining  the  middle  points  of  any  two  sides  is  parallel  to  the  third 
side  and  equal  to  one  half  of  it. 

18.  If  (2,  1),  (3,  3),  (6,  2)  are  the  middle  points  of  the  sides  of  a  triangle, 
what  are  the  coordinates  of  the  vertices  of  the  triangle  ? 

Ans.    (-1,2),  (5,0),  (7,4). 

19.  Three  vertices  of  a  parallelogram  are  (1,  2),  (-5,  —3),  (7,-6). 
What  are  the  coordinates  of  the  fourth  vertex  ? 

Ans.    (1,  -  11),  (-  11,  5),  or  (13,  -  1). 

20.  The  middle  point  of  a  line  is  (6,  4),  and  one  end  of  the  line  is  (5,  7). 
What  are  the  coordinates  of  tlie  other  end  ?  Ans.    (7,  1). 

21.  The  vertices  of  a  triangle  are  (2,  3),  (4,  -  5),  (-  3,  -  6).  Find  the 
coordinates  of  the  point  where  the  medians  intersect  (center  of  gravity). 


42  ANALYTIC  GEOMETRY 

22.  Find  the  area  of  the  isosceles  triangle  whose  vertices  are  (1,  5),  (5,  1), 
(—9,  —  9)  by  finding  the  lengths  of  the  base  and  altitude, 

23.  A  line  AB  is  produced  to  C  so  that  BC  =  |  AB.  If  the  points  A  and  B 
have  the  coordinates  (5,  6)  and  (7,  2)  respectively,  what  are  the  coordinates 
of  C  ?  Ans.    (8,  0). 

24.  Show  that  formula  (VII)  holds  for  oblique  coordinates,  that  is,  Z  XOY 
may  have  any  value. 

25.  How  far  is  the  point  bisecting  the  line  joining  the  points  (5,  5)  and  (8,  7) 
from  the  origin  ?     What  is  the  slope  of  this  last  line  ?  Ans.    2  Vl3,  |. 

24.  Areas.  In  this  section  the  problem  of  determining  the  area 
of  any  polygon  the  coordinates  of  whose  vertices  are  given  will 
be  solved.     We  begin  with 

Theorem  VIII.  The  area  of  a  triangle  whose  vertices  are  the 
origin,  P^  (a^i,  y-^,  and  Pc,  (x^,  y^  is  give7i  by  the  formula 

(VIII)         Area  of  triangle  OP^P^  =  i(oo^y2  —  ^22/i)- 

f.fe2/J  ■^''•''•^*    In  the  figure  let 


a  = 

=  AXOP„ 

P- 

--  Z  XOP^, 

e^ 

--  Z  P,OP^. 

^  = 

=  y8-a. 

3fa  My    JL 

(1) 

By  18,  p.  20, 

(2)  Area  A  OP^P^  =  ^0P^-  OP^  sin  0 

=  \0P^-  OP,  sin (yS  -  a)  [by  (1)] 

(3)  =  \  OPi  •  OP2  (sin  ft  cos  a  —  cos  ft  sin  or). 

But  m  the  figure 

sm  ft  =  -^-^  =  -^^^j  cos  B  = =  — ^> 

^        OP,        OP,  '^       OP,       OP, 

M^P^        y^  OM.         X. 

sm  a  =  — ^— ^  =  -^^-^,  cos  a  = =  — - • 

OPi        OPi  OPi        OP^ 

Substituting  in  (3)  and  reducing,  we  obtain 

Area  A  OP^P,  =  ^  {x^y,  -  x,y^.  q.e.d. 


CARTESIAN  COORDINATES 


43 


Ex.  1.    Find  the  area  of  the  triangle  whose  vertices  are  the  origin,  (—2,  4), 
and  (-  5,  -  1). 

Solution.    Denote  ( -  2,  4)  by  Pi,  ( -  5,  - 1)  by  Pg. 
Then 

cci  =  -  2,  2/1  =  4,  X2  =  -  5,  2/2  =  -  1- 


(-\i) 

YJk 

/ 

\ 

f 

\ 

\/ 

\ 

a,i) 

y 

\ 

I 

^. 

^ 

X 

r-s 

-t) 

Substituting  in  (VIII), 

Area  =  i  [-  2  •  -  1  -  (-  5)  •  4]  =  11. 
Then  Area  =  11  unit  squares. 

If,  however,  the  formula  (VIII)  is  applied  by  denoting  (—2,  4)  by  P2,  and 
(_  5^  -  1)  by  Pi,  the  result  will  be  -  11. 
The  two  figures  are  as  follows  : 


(1)  (2) 

The  cases  of  positive  and  negative  area  are  distinguished  by 
Theorem  IX.    Passing  around  the  perimeter  in  the  order  of  the 
vertices  0,  P^,  P^, 

if  the  area  is  on  the  left,  as  in  Fig.  1,  then  (VIII)  gives  a  posi- 
tive restdt; 
if  the  area  is  on  the  right,  as  in  Fig.  2,  then  (YIII)  gives  a 
negative  result. 

Proof  When  the  area  is  on  the  left  as  in  Fig.  1,  then  in  (1), 
p.  42,  we  have  fi  >  a,  and  hence  0  is  positive.  Therefore  sin  $ 
is   positive  and   the  product   in   (2),      „  p 

p.  42,  which  gives  the  area  of  OP^P^, 
is  also  positive.  But  when  the  area 
is  on  the  right,  as  in  Fig.  2,  we  have 
13  <  a,  and  hence  6  is  negative.  Then 
sin  0  is  negative,  and  hence  also  the  product  in  (2),  p.  42,  which 
gives  the  area  of  OP^P^.  q.e.d. 

Formula  (VIII)  is  easily  applied  to  any  polygon  by  regarding 
its  area  as  made  up  of  triangles  with  the  origin  as  a  common 
vertex.    Consider  any  triangle. 


.^1 


.e+ 


(1) 


(2) 


44 


ANALYTIC  GEOMETRY 


Theorem  X.  The  area  of  a  triangle  whose  vertices  are  P^  (x-^,  yi), 
A(^2j  2/2);  ^3(^3,  2/3)  ««  ff^^en  by 

(X)  Area  A  1*1  jP2jP3=i  (a:5i2/2  —  a?22/i+^2?^3—»^32/2 +^32/1  —  ^12/3)- 
This  formula  gives  a  positive  or  negative  result  according  as  the 
area  lies  to  the  left  or  right  in  passing 
around  the  perimeter  in  the  order  P^P^Pz- 
Proof  Two  cases  must  be  distin- 
guished according  as  the  origin  is 
within  or  without  the  triangle. 

Fig.    1,    origin   within    the    triangle. 
By  inspection, 

(5)  Area  A  P^P^P^  =  A  OP^P^  +  A  OP^P.^  +  A  OP^P^, 
since  these  areas  all  have  the  same  sign.  1  p 

yig.  2,  origin  without  the  triangle.     By  inspection, 

(6)  Area  A  P^P^P^  =  A  OP.P^  +  A  OP,P,  +  A  OP,P„ 

since  OP^P^,  OP^Pi  have  the  same  sign,  but  OP^P-s  the  opposite 
sign,  the  algebraic  sum  giving  the  desired  area. 

By  (VIII),  A  OP,P,  =  ^(x,7/,  -  X0,), 

A  OP^P^  =  ^  (x^y^  -  x^y^),  A  OP^P^  =  \  (x^y^  -  x^y^). 

Substituting  in  (5)  and  (6),  we  have  (X). 

Also  in  (5)  the  area  is  positive,  in  (6)  negative. 

An  easy  way  to  apply  (X)  is  given  by  the  following 

Rule  for  finding  the  area  of  a  triangle. 

First  step.  Write  down  the  vertices  in  two  columns, 
abscissas  in  one,  ordinates  in  the  other,  repeating  the 
coordinates  of  the  first  vertex. 

Second  step.  Multiply  each  abscissa  by  the  ordinate  of  the  next 
row,  and  add  results.      This  gives  x^y^  4-  ^^yz  +  ^zVi- 

Third  step.  Multiply  each  ordinate  by  the  abscissa  of  the  next 
row,  and  add  residts.      This  gives  y^x^  +  2/2^3  +  Vz^x- 

Fourth  step.  Subtract  the  result  of  the  third  step  from  that  of 
the  second  step,  and  divide  by  2.  This  gives  the  required  area, 
namely,  formula  (X). 


Q.E.D. 


Xi 

•2/1 

X2 

Vi 

a^3 

Vz 

Xi 

7 

y\ 

CARTESIAN  COORDINATES 


45 


-2  % 


It  is  easy  to  show  in  the  same  manner  that  the  rule  applies  to 
any  polygon,  if  the  following  caution  be  observed  in  the  first  step  : 

Write  down  the  coordinates  of  the  vertices  in  an  order  agreeing 
with  that  established  by  passing  continuously  around  the  perimeter , 
and  repeat  the  coordinates  of  the  first  vertex. 

Ex.  2.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (1,  6), 
(-3,  -4),  (2,  -2),  (-1,3). 

Solution.    Plotting,  we  have  the  figure  from  which 
we  choose  the  order  of  the  vertices  as  indi-         ..  ^ 

cated  by  the  arrows.    Following  the  rule  :     _  i         3 

First  step.    Write  down  the  vertices  in     —  3     —  4 
order.  2     —  2 

Second  step.    Multiply   each    abscissa 
by  the  ordinate  of  the  next  row,  and  add.    This  gives 

lx3  +  (-lx-4)  +  (-3x-2)  +  2x6=3  25. 

Third  step.    Multiply  each  ordinate  by  the  abscissa 
of  the  next  row  and  add.     This  gives 
6x-l  +  3x-3  +  (-4x2)  +  (-2xl)=-25. 

Fourth  step.    Subtract  the  result  of  the  third  step 

from  the  result  of  the  second  step,  and  divide  by  2. 

.            25  4-25      ^„       .^  . 

.-.  Area  = =  25  unit  squares.     Ans. 

The  result  has  the  positive  sign,  since  the  area  is  on  the  left. 


PROBLEMS 

1.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3),  (1,  5),  (—  1,  —  2). 

Ans.   y. 

2.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3),  (4,  —5),  (  —  3,  —6). 

Ans.  29. 

3.  Find  the  area  of  the  triangle  whose  vertices  are  (8,  3),  (—  2,  3),  (4,  —  5). 

Ans.  40. 

4.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  0),  (—  a,  0),  (0,  b). 

Ans.  ah. 

5.  Find  the  area  of  the  triangle  whose  vertices  are  (0,  0),  (xi,  yi),  (X2, 2/2)- 

Ans.  ^jyiH^^. 


0^ 
46  ANALYTIC  GEOMETRY 

6.  Find  the  area  of  the  triangler^whose  vertices  are  (a,  1),  (0,  6),  (c,  1). 

2 

7.  Eind  the  area  of  the  triangle  whose  vertices  are  (a,  6),  (&,  a),  (c,  -  c). 

8.  Find  the  area  of  the  triangle  whose  vertices  are  (3, 0) ,  (0, 3  V3),  (6, 3  V3). 

An8.  9V3. 

9.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(2,  3),  (5,  4),  (—4,  1)  is  zero,  and  hence  that  these  points  all  lie  on  the  same 
straight  line. 

10.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(a,  6  +  c),  (&,  c  ■\-  a)^  (c,  a  +  6)  is  zero,  and  hence  that  these  points  all  lie  on 
the  same  straight  line. 

11.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(a,  c  +  a),  (—  c,  0),  (—a,  c  —  a)  is  zero,  and  hence  that  these  points  all  lie 
on  the  same  straight  line. 

12.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (-2,  3), 
(-3,   -4),   (5,    -1),   (2,  2).  Ans.  31. 

13.  Find  the  area  of  the  pentagon  whose  vertices  are  (1,  2),  (3,  —  1), 
(6,  -  2),  (2,  5),  (4,  4).  Ans.   18. 

14.  Find  the  area  of  the  parallelogram  whose  vertices  are  (10,  5),  (—  2,  5), 
(-  5,  -  3),  (7,  -  3).  Ans.  96. 

15.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (0,  0),  (5,  0), 
(9,  11),  (0,  3).  Ans.  41. 

16.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (7,  0),-  (11,  9), 
(0,  5),  (0,  0).  Ans.  59. 

17.  Show  that  the  area  of  the  triangle  whose  vertices  are  (4,  6),  (2,  —  4), 
(—4,  2)  is  four  times  the  area  of  the  triangle  formed  by  joining  the  middle 
points  of  the  sides. 

18.  Show  that  the  lines  drawn  from  the  vertices  (3,  —  8),  (—  4,  6),  (7,  0) 
to  the  medial  point  of  the  triangle  divide  it  into  three  triangles  of  equal  area. 

19.  Given  the  quadrilateral  whose  vertices  are  (0,  0),  (6,  8),  (10,  —  2), 
(4,  —  4) ;  show  that  the  area  of  the  quadrilateral  formed  by  joining  the 
middle  points  of  its  adjacent  sides  is  equal  to  one  half  the  area  of  the  given 
quadrilateral. 


CARTESIAN  COORDINATES  47 

25.  Second  theorem  of  projectiqj|j 

Lemma  I.    If  M^^  M^,  M^  are  any  imree  points  on  a  directed  line, 
then  in  all  cases 


T 


Jfi 


M-i 


M, 


Mo 


M^ 


0  Ml 

Proof.    Let  0  be  the  origin. 

By  (1),  p.  24,  M^M^  =  OM^  -  OM^, 

M^Ms  =  OMs  -  OM^. 

Adding,  M^M^  +  M^M^  =  OM^  -  OM^. 

But  by  (1),  p.  24,         M^M^  =  OM^  -  OM^. 


Q.E.D. 


This  result  is  easily  extended  to  prove 

Lemma  II.    If  Mi,  M^,  M^,  •••,  ikf„_i,  M^  are  any  n  points  on  a 
directed  line,  then  in  all  cases 

MiM^  =  M1M2  -h  M^M^  +  M^M^  +  •  •  •  +  M„_,M„, 

the  lengths  in  the  right-hand  member  being  so  written  that  the 
second  point  of  each  length  is  the  first  point  of  the  next. 

The  line  joining  the  first  and  last  points  of  a  broken  line  is 
called  the  closing  line. 


Cii/,  M>, 


M^D 


Thus  in  Fig.  1  the  closing  line  is  PyP^ ;  in  Fig.  2  the  closing 
line  is  P1P5. 


48 


ANALYTIC  GEOMETRY 


Theorem  XI.  Second  theorem  of  projection.  If  each  segment  of  a 
broken  line  be  given  the  direction  determined  in  passing  continuously 
from  one  extremity  to  the  other,  then  the  algebraic  sum  of  the  pro- 
jections of  the  segments  upon  any  directed  line  equals  the  projection 
of  the  closing  line. 


Proof 
in  Fig.  1 


The  proof  results  immediately  from  the  Lemmas.    For 

M1M2  =  projection  of  P1P2  j 
M^M^  =  projection  of  P2P3  5 
MiM^  =  projection  of  closing  line  PiPs- 

But  by  Lemma  I 

Ml  Mo  +  M^M^=  M^Ms, 
and  the  theorem  follows.  ^ 

Similarly  in  Fig.  2.  q.e.d. 

Corollary.  If  the  sides  of  a  closed  polygon  be  given  the  direction 
established  by  passing  continuously  around  the  perimeter,  the  sum 
of  the  projections  of  the  sides  upon  any  directed  line  is  zero. 

For  tlje  closing  line  is  now  zero. 

Ex.  1.    Find  the  projection  of  the  line  joining  the  origin  and  (5,  3)  upon 

a  line  passing  through  ( —  5,  0)  whose 

inclination  is  —  • 
4 

Solution.    In  the  figure,  applying  the 
second  theorem  of  projection, 

proj.  of  OP  on  AB 

=  proj.  of  OM  4-  proj.  of  MP 

It  It 

=  OM  cos  —  +  MP  cos  — 

4  4 

(by  first  theorem  of  projection,  p,  30) 

=  5V2-i-fV2  =  4V2.    Ans. 

The  essential  point  in  the  solution  of  problems  like  Ex.  1  is  the  replacing 
of  the  given  line,  by  means  of  Theorem  XI,  by  a  broken  line  with  two  seg- 
ments which  are  parallel  to  the  axes. 


I 


CARTESIAN  COORDINATES 


49 


Ex.  2.    Find  the  perpendicular  distance  from  the  Ime  passing  through 
(4,  0),  whose  inclination  is  — ,  to  the 
point  (10,  2).  ^ 

Solution.    In  the  figure  draw  OC 
perpendicular  to  the  given  line  AB. 

27t 


ZXAS 


or  120°. 


.-.  Z  XOS  =  30°,  Z  SOY  =  60°. 

Required  the  perpendicular  dis- 
tance RP. 

Project  the  broken  line  OMP  upon 
OC.  Then,  by  the  second  theorem  of 
projection, 


(1) 


But  in  the  figure 


(2) 


proj.  of  OP  =  proj.  of  OM  +  proj.  of  MP 

=  OM  cos  Z  XOS  +  MP  cos  ZSOY 
=  10.iV3  +  2.i 
=  1  +  5  Vs. 

proj.  of  OP  =  05 +  -87 

=  OA  cos^XOS  +  RP 
=  4  .  i  Vs  +  RP. 


From  (1)  and  (2), 


EP  +  2V3  =  1  +  5V3. 

RP  =  l-\-SVs.     Ans. 


PROBLEMS 

1.  Four  points  lie  on  the  axis  of  abscissas  at  distances  of  1,  3,  6,  and  10 
respectively  from  the  origin.     Find  P1P4  by  Lemma  II. 

2.  A  broken  line  joins  continuously  the  points  (—1,  4),  (3,  6),  (6,  —  2), 
(8,  1),  (1,  —  1).  Show  that  the  second  theorem  of  projection  holds  when  the 
segments  are  projected  on  the  X-axis. 

3.  Show  by  means  of  a  figure  that  the  projection  of  the  broken  line  join- 
[    ing  the  points  (1,  2),  (5,  4),  (-  1,  -  4),  (3,  -  1),  and  (1,  2)  upon  any  line  is 

zero. 

4.  Find  the  projection  of  the  line  joining  the  points  (2,  1)  and  (5,  3)  upon 

7t 


a  line  passing  through  the  point  (—1,  1)  whose  inclination  is 


Ans. 


3  V3  +  2 


50  ANALYTIC  GEOMETRY 

5.  What  is  the  projection  of  the  line  joining  these  same  points  upon  any 
line  whose  inclination  is  ^  ?     Why  ? 

6.  Find  the  projection  of  the  line  joining  the  points  (-1,  3)  and  (2,  4) 
upon  any  line  whose  inclination  is  f  it.  Ans.   —  V 2. 

7.  Find  the  projection  of  the  broken  line  joining  the  points  (-  1,  4), 

It 
(3,  6),  and  (5,  0)  upon  a  line  whose  inclination  is  -•     Verify  your  result  by 

finding  the  projection  of  the  closing  line.  Au8.   V 2, 

8.  Find  the  projection  of  the  broken  line-joining  (0,  0),  (4,  2),  and  (6,  -  3) 

upon  a  line  whose  inclination  is  -r-  •  Ans.   — — — 

o  2 

9.  Show  that  the  projection  of  the  sides  of  the  triangle  (2,  1),  (-  1,  5), 
(—3,  1)  upon  a  line  whose  inclination  is  —  is  zero. 

10.  Find  the  perpendicular  distance  from  the  point  (6,  3)  to  a  line  passing 

It  7 

through  the  point  (—4,  0)  with  an  inclination  of  —•  Ans.   — p- 

11.  Find  the  perpendicular  distance  from  the  point  (—5,  —  1)  to  a  line 
passing  through  the  point  (6,  0)  and  having  an  inclination  of  f  it. 

Anjs.   6V2. 

12.  A  line  of  inclination  —  passes  through  the  point  (5,  0).    Find  the  per- 
pendicular distance  to  the  parallel  line  passing  through  the  point  (0,  2). 

5  +  2  V3 
Ans. 


CHAPTER  III 
THE  CURVE  AND  THE  EQUATION 

26.  Locus  of  a  point  satisfying  a  given  condition.  The  curve* 
(or  group  of  curves)  passing  through  'all  points  which  satisfy  a 
given  condition,  and  through  no  other  points,  is  called  the  locus 
of  the  point  satisfying  that  condition. 

For  example,  in  Plane  Geometry,  the  following  results  are 
proved  : 

The  perpendicular  bisector  of  the  line  joining  two  fixed  points 
is  the  locus  of  all  points  equidistant  from  these  points. 

The  bisectors  of  the  adjacent  angles  formed  by  two  lines  is 
the  locus  of  all  points  equidistant  from  these  lines. 

To  solve  any  locus  problem  involves  two  things : 

1.  To  draw  the  locus  by  constructing  a  sufficient  number  of 
points  satisfying  the  given  condition  and  therefore  lying  on  the 
locus. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve,  t 

Analytic  Geometry  is  peculiarly  adapted  to  the  solution  of 
both  parts  of  a  locus  problem. 

27.  Equation  of  the  locus  of  a  point  satisfying  a  given 
condition.  Let  us  take  up  the  locus  problem,  making  use  of  coor- 
dinates. If  any  point  P  satisfying  the  given  condition  and  there- 
fore lying  on  the  locus  be  given  the  coordinates  (x,  y),  then  the 
given  condition  will  lead  to  an  equation  involving  the  variables 
X  and  y.  The  following  example  illustrates  this  fact,  which  is 
of  fundamental  importance. 

*  The  word  "  curve"  wiU  hereafter  signify  any  contimious  line,  straight  or  curved. 

t  As  the  only  loci  considered  in  Elementary  Geometry  are  straight  lines  and  circles, 
the  complete  loci  may  be  constructed  by  ruler  and  compasses,  and  the  second  part  is 
relatively  unimportant. 

51    '-^ 


52 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  equation  in  x  and  y  if  the  point  whose  locus  is  required 
shall  be  equidistant  from  A{—  2,  0)  and  B{—  3,  8). 

Solution.    Let  P  (x,  y)  be  any  point  on  the  locus.  Then  by  the  given  condition 
(1)  PA  =  PB. 

But,  by  formula  IV,  p.  31, 


PA  =  V{x  +  2)^  +  {y- 

-0)2, 

P5=V(a:  +  3)2  +  (?/- 
Substituting  in  (1), 

-8)2. 

V(x  +  2)'^  +  (2/  -  0)2 

(2)  

=  V(x  +  3)2  +  {y-  8)2. 

Squaring  and  reducing, 

(3)  2x  -  162/ -h  69  =  0. 

In  the  equation  (3),  x  and  y  are  variables  representing  the  coordinates  of 
any  point  on  the  locus,  that  is,  of  any  point  on  the  perpendicular  bisector  of 
the  line  AB.     This  equation  has  two  important  and  characteristic  properties : 

1.  The  coordinates  of  any  point  on  the  locus  may  be  substituted  for  x 
and  y  in  the  equation  (3),  and  the  result  will  be  true. 

For  let  Pi  (xi,  yi)  be  any  point  on  the  locus.  Then  PiA  =  PiB,  by  defi- 
nition.    Hence,  by  formula  IV,  p.  31, 

(4)  V(xi  +  2)2  +  2/i2  =  V(xi  +  3)2  +  (^1  -  8)2, 
or,  squaring  and  reducing, 

(5)  2  xi  -  16  yi  +  69  =  0. 
Therefore  Xi  and  yi  satisfy  (3). 

2.  Conversely,  every  point  whose  coordinates  satisfy  (3)  will  lie  upon  the 
locus. 

For  if  Pi(Xi,  yi)  is  a  point  whose  coordinates  satisfy  (3),  then  (5)  is  true, 
and  hence  also  (4)  holds.  q.e.d. 

In  particular,  the  coordinates  of  the  middle  point  C  ot  A  and  B,  namely, 
x=-2i,y=:4:  (Corollary,  p.  39),  satisfy  (3),  since  2 ( -  2i) - 16  x  4  +  69  =  0. 

This  example  illustrates  the  following  correspo?idence  between 
Pure  and  Analytic  Geometry  as  regards  the  locus  problem  : 

Locus  problem 
Pure  Geometry  Analytic  Geometry 

The  geometrical  condition  (satis-      An  equation  in  the  variables  x 
fied  by  every  point  on  the  and?/ representing  coordinates 

locus).  (satisfied  by  the  coordinates 

of  every  point  on  the  locus). 


THE  CURVE  AND  THE  EQUATION 


53 


This  discussion  leads  to  the  fundamental  definition : 

The  equation  of  the  locus  of  a  point  satisfying  a  given  condition 
is  an  equation  in  the  variables  x  and  y  representing  coordinates 
such  that  (1)  the  coordinates  of  every  point  on  the  locus  will 
satisfy  the  equation;  and  (2)  conversely,  every  point  whose 
coordinates  satisfy  the  equation  will  lie  upon  the  locus.  • 

This  definition  shows  that  the  equation  of  the  locus  must  be 
tested  in  two  ways  after  derivation,  as  illustrated  in  the  example 
of  this  section  and  in  those  following. 

From  the  above  definition  follows  at  once  the 

Corollary.  A  point  lies  upon  a  cun^e  when  and  only  when  its 
coordinates  satisfy  the  equation  of  the  curve. 

28.  First  fundamental  problem.  To  find  the  equation  of  a 
curve  which  is  defined  as  the  locus  of  a  point  satisfying  a  giuen 
condition. , 

The  following  rule  will  suffice  for  the  solution  of  this  problem 
in  many  cases : 

Rule.  First  step.  Assume  that  P  (x,  y)  is  any  point  satisfying 
the  given  condition  and  is  therefore  on  the  curve. 

Second  step.     Write  down  the  given  condition. 

Third  step.  Express  the .  given  condition  in  coordinates  and 
simplify  the  result.  The  final  equation,  containing  x,  y,  and  the' 
given  constants  of  the  problem,  will  be  the  required  equation. 

Ex.  1.    Find  the  equation  of  the  straight  line  passing  through  Pi  (4,  —  1) 

3  7t 

and  having  an  inclination  of 

Solution.    First  step.    Assume  P{x,  y)  any  point 
on  the  line. 

Second  step.    The  given  condition,  since  the  incli- 

•    Stt 
nation  a  is  —  ,  may  be  written 

4 
(1)  Slope  of  PiP  =  tan  a  =  -  1. 

Third  step.    From  (V),  p.  35, 


.N 

\ 

/_ 

1 

1 

\ 

-.</ 

) 

s 

0 

r4, 

I) 

X 

j2 

\ 

(2) 


Slope  of  PiP  =  tan  a  = 


yi-Vi      2/  +  1 


Xi  —  X2        X  —  4 
[By  substituting  {x,  y)  for  {x^,  y^),  and  (4,  —  1)  for  (Xj,  y,).] 


54 


ANALYTIC  GEOMETKY 


/.  from  (1), 


or 
(3) 


2/  +  1  1 

X  —  4 

x-\-  y  —  S  =  0.     Ans. 


To  prove  that  (3)  is  the  required  equation  : 

1.  The  coordinates  (xi,  yi)  of  any  point  on  the  line  will  satisfy  (3),  for 
the  line  joins  (xi,  yi)  and  (4,  —  1),  and  its  slope  is  —  1 ;  hence,  by  (V),  p.  35, 
substituting  (4,  —  1)  for  (X2,  ^2), 


-1 


yi  +  1 

Xi  —  4 


or  Xi  +  ?/i  -  3  =  0, 


and  therefore  Xi  and  2/1  satisfy  the  equation  (3). 

2,  Conversely,  any  point  whose  coordinates  satisfy  (3)  is  a  point  on  the 
straight  line.     For  if  (xi,  2/1)  is  any  such  point,  that  is,  if  Xi  +  2/1  —  3  =  0,  then 
2/1  +  1 


also 


is  true,  and  (xi,  2/1)  is  a  point  on  the  line  passing  through 


(4,  —  1)  and  having  an  inclination  equal  to  —7-  ■ 


Q.E.D. 


Ex.  2.    Find  the  equation  of  a  straight  line  parallel  to  the  axis  of  y  and 
at  a  distance  of  6  units  to  the  right. 

Solution.    First  step.    Assume  that  P(x,  y)  is 
any  point  on  the  line,  and  draw  NP  perpendicular 
to  OF. 
^^)        Second   step.     The  given  condition  may   be 
written 

(4)  NP  =  6. 


Yi 

1^ 

N 

P 

^ 

(^ 

0 

M 

Third  step.    Since  NP  =  OM  =  x,  (4)  becomes 
(5)  X  =  6.     Ans. 


The  equation  (5)  is  the  required  equation  : 

1.  The  coordinates  of  every  point  satisfying  the  given  condition  may  be 
substituted  in  (5).  For  if  Pi(xi,  2/1)  is  any  such  point,  then  by  the  given 
condition  Xi  =  6,  that  is,  (xi,  2/1)  satisfies  (5). 

2.  Conversely,  if  the  coordinates  (xi,  2/1)  satisfy  (5),  then  Xi  =  6,  and 
Pi{^u  Vi)  is  at  a  distance  of  six  units  to  the  right  of  YY^.  q.e.d. 

The  method-  above  illustrated  of  proving  that  the  derived  equation  has  the 
two  characteristic  properties  of  the  equation  of  the  locus  should  be  carefully 
studied  and  applied  to  each  of  the  following  examples. 


THE  CURVE  AND  THE  EQUATION 


55 


V(x+  1)2  +  (y  -  2)2. 
Substituting  in  (6), 

V(x  +  1)2  +  (y  -  2)2  =  4. 
Squaring  and  reducing, 


i% 


Ex.  3.    Find  tlie  equation  of  tlie  ^ocus  of  a  point  whose  distance  from 
(—1,  2)  is  always  equal  to  4. 

Solution.  First  step.  Assume  that  P(aj,2/) 
is  any  point  on  the  locus. 

Second  step.    Denoting  (—1,  2)  by  C, 
the  given  condition  is 
(6)  ■  PC  =  4. 

Third  step.    By  formula  (IV),  p.  31, 
PC 


_I!^'^? 


■U) 


(7) 


a;2  +  y2  +  2x-42/-ll  =  0. 


This  is  the  required  equation,  namely,  the  equation  of  the  circle  whose 
center  is  (—1,  2)  and  radius  equals  4.  The  method  of  proof  is  the  same 
as  that  of  the  preceding  examples. 


PROBLEMS 

1.  Find  the  equation  of  a  line  parallel  to  OF  and 

(a)  at  a  distance  of  4  units  to  the  right. 

(b)  at  a  distance  of  7  units  to  the  left. 

(c)  at  a  distance  of  2  units  to  the  right  of  (3,  2). 

(d)  at  a  distance  of  5  units  to  the  left  of  (2,  —  2). 

2.  What  is  the  equation  of  a  line  parallel  to  OF  and  a  —  b  units  from  it  ? 
How  does  this  line  lie  relative  to  OF  if  a>6>0?   ifO>6>a? 

3.  Find  the  equation  of  a  line  parallel  to  OX  and 

(a)  at  a  distance  of  3  units  above  OX. 

(b)  at  a  distance  of  6  units  below  OX. 

(c)  at  a  distance  of  7  units  above  (—  2,  —  3). 

(d)  at  a  distance  of  5  units  below  (4,  —  2). 

4.  What  is  the  equation  of  XX?  of  YY'? 

6.  Find  the  equation  of  a  line  parallel  to  the  line  x  =  4  and  3  units  to  the 
right  of  it.     Eight  units  to  the  left  of  it. 

6.  Find  the  equation  of  a  line  parallel  to  the  line  y  =  -2  and  4  units 
below  it.     Five  units  above  it. 

7.  How  does  the  line  y  =  a-61ieifa>6>0?  if6>a>0? 

8.  What  is  the  equation  of  the  axis  of  x  ?  of  the  axis  oly? 


(c)  Pi  is  (-  2,  3)  and  7n  = ?  Ans.  V2x-2y-\-6+2V2  =  0. 


3^ 
5Q    "         '  ANALYTIC   GEOMETKY 

9.  What  is  the  equation  of  the  locus  of  a  point  which  moves  always  at 
a  distance  of  2  units  from  the  axis  of  x  ?  from  the  axis  of  y?  from  the  line 
X  =—  6?  from  the  line  y  =  4? 

10.  What  is  the  equation  of  the  locus  of  a  point  which  moves  so  as  to 
be  equidistant  from  the  lines  x  =  b  and  x  =  9?  equidistant  from  y  =  S  and 
y=-7? 

11.  What  are  the  equations  of  the  sides  of  the  rectangle  whose  vertices 
are  (5,  2),  (5,  5),  (- 2,  2),  (- 2,  5)'? 

In  problems  12  and  13,  Pi  is  a  given  point  on  the  required  line,  m  is  the 
slope  of  the  line,  and  a  its  inclination. 

12.  What  is  the  equation  of  a  line  if 

(a)  Pi  is  (0,  3)  and  ?n  =  -  3  ?  Ans.  3x  +  ?/-3  =  0. 

(b)  Pi  is  (-  4,  -  2)  and  m  =  i?  Ans.  x-  ?jy  -2  =  0. 

2 

V3 

(d)  Pi  is  (0,  5)  and  m  =:  ^?  Ans.  V3x-2?/  +  10  =  0. 

(e)  Pi  is  (0,  0)  and  m  1:=  -  I  ?  Ans.  2x  +  Sy  =  0. 

(f)  Pi  is  (a,  b)  and  m  =  0  ?  Ans.  y  =  b. 

(g)  Pi  is  (—  a,  b)  and  ni  =  'X)?  Ans.  x  =—  a. 

13.  What  is  the  equation  of  a  line  if 

(a)  Pi  is  (2,  3)  and  a  =  45°  ?  Ans.  x  -  y  -\-  1  =  0. 

(b)  Pi  is  (-  1,  2)  and  a  =  45° ?  Ans.  x  -  y  -]- S  =  0. 

(c)  Pi  is  ( -  a,  -  b)  and  a:  =  45°  ?  Ans.  x  -  y  =  b  -  a. 

(d)  Pi  is  (5,  2)  and  a  =  60°?  Ans.   Vsx  -  y -^  2  -  ^Vs  =  0. 

(e)  Pi  is  (0,  -  7)  and  or  =  60°  ?  Ans.   -VSx-y  -7  =  0. 

(f)  Pi  is  (-  4,  5)  and  a  =  0°?  Ans.  y  =  6. 

(g)  Pi  is  (2,  -  3)  and  a  =  90°?  Ans.  x  =  2. 
(h)  Pi  is  (3,  -  3  V3)  and  a:  =  120° ?  Ans.   ^Sx  +  y  =  0. 

(i)  Pi 'is  (0,  3)  and  a  =  150°?  Ans.   Vsx  -\-Sy  -9  =  0. 

(j)  Pi  is  (a,  6)  and  a  =  135°  ?  Ans.  x  -{-  y  =  a  -\-  b. 

14.  Are  the  points  (3,  9),  (4,  6),  (5,  5)  on  the  line  3x  +  2y  =  25? 

15.  Find  the  equation  of  the  circle  with 

(a)  center  at  (3,  2)  and  radius  =  4.  Ans.  x^  +  y^  -  6 x  —  4y  -  S  =  0. 

(b)  center  at  (12,  -  5)  and  r  =  13.  Ans.  x^  +  y^  -  24x -\-10y  =  0. 

(c)  center  at  (0,  0)  and  radius  =  r.  Ans.  x^  -\-  7/'^  =  r^. 

(d)  center  at  (0,  0)  and  r  =  5.  Ans.  x'^  +  7/^  =  25. 

(e)  center  at  (3  a,  4  a)  and  r  =  5  a.  Ans.  x^  +  y^  _  2  a  (3  cc  +  4  y)  =  0. 

(f)  center  at  (6  +  c,  6  —  c)  and  r  =  c. 
Ans.  x2  +  ?/2  -  2  (&  +  c)  X  -  2  (6  -  c)  ?/  +  2  62  4.  c2  =  0. 


THE  CURVE  AND  THE  EQUATION         67 

16.  Find  the  equation  of  a  circle  whose  center  is  (5,  —  4)  and  whose 
circumference  passes  through  the  point  (—2,  3). 

17.  Find  the  equation  of  a  circle  having  the  line  joining  (3,  —  5)  and 
(—  2,  2)  as  a  diameter. 

'    18.  Find  the  equation  of  a  circle  touching  each  axis  at  a  distance  6  units 
from  the  origin. 

19.  Find  the  equation  of  a  circle  whose  center  is  the  middle  point  of  the 
line  joining  (—6,  8)  to  the  origin  and  whose  circumference  passes  through 
the  point  (2,  3). 

20.  A  point  moves  so  that  its  distances  from  the  two  fixed  points  (2,  —  3) 
and  (—1,  4)  are  equal.     Find  the  equation  of  the  locus  and  plot. 

Ans.   Sx-7ij  +  2  =  0. 

21.  Find  the  equation  of  the  perpendicular  bisector  of  the  line  joining 

(a)  (2,  1),  (-  3,  -  3).  Ans.  lOx  +  8y -\- IS  =  0 

(b)  (3,  1),  (2,  4).  Ans.  x  -  3 y  +  5  =  0. 

(c)  (-1,  -1),  (3,  7).  Ans.  x-\-2y-7  =  0. 

(d)  (0,  4),  (3,  0).  Ans.  6x-Sy-\-7  =0. 

(e)  (xi,  vi),  {X2,  2/2). 

Ans.  2  (xi  -  X2)  X  +  2  (yi  -  2/2)  2/  +  X2^  -  x^  +  y^^  -  y^  =  0. 

22.  Show  that  in  problem  21  the  coordinates  of  the  middle  point  of 
the  line  joining  the  given  points  satisfy  the  equation  of  the  perpendicular 
bisector. 

23.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of  the 
triangle  (4,  8),  (10,  0),  (6,  2).     Show  that  they  meet  in  the  point  (11,  7). 

24^  Express  by  an  equation  that  the  point  (h,  k)  is  equidistant  from 
(-  1,  1)  and  (1,  2) ;  also  from  (1,  2)  and  (1,  -  2).  Then  show  that  the  point 
(f,  0)  is  equidistant  from  (-  1,  1),  (1,  2),  (1,  -  2). 

29.  General  equations  of  the  straight  line  and  circle.     The 

methods  illustrated  in  the  preceding  section  enable  us  to  state 
the  following  results : 

1.  A  straight  line  parallel  to  the  axis  of  y  has  an  equation  of 
the  form  x  =  constant. 

2.  A  straight  line  parallel  to  the  axis  of  x  has  an  equation  of 
the  form  y  =  constant. 


58  ANALYTIC  GEOMETRY 

Theorem  I.    The  equation  of  the  straight  line  passing  through  a 
point  B  (0,  h)  on  the  axis  of  y  and  having  its  slope  equal  to  m  is 

(I)  y  =  fiidc  +  h. 

Proof    First  step.    Assume  that  P  (x,  y)  is  any  point  on  the  line. 
Second  step.    The  given  condition  may  be  written 

Slope  of  PB==m. 
Third  step.    Since  by  Theorem  V,  p.  35, 

Slopeof  Pi5  =  ^^^, 

[Substituting  (x,  y)  for  (xi,  yi)  and  (0,  6)  for  {X2,  2/2)] 

^1  y  —  h 

then  =  m,  or  2/  =  mx  +  b.  q.e.d. 

Theorem  II.    The  equation  of  the  circle  whose  center  is  a  given 
point  (a,  (3)  and  whose  radius  equals  r  is 

(II)  00^  +  y^  -2aiic-2py  +  a^  -\-  P^  -r^  =  0. 

Proof    First  step.    Assume  that  P  (x,  y)  is  any  point  on  the 

locus. 

Second  step.    If  the  center  (a,  /?)  be  denoted  by  C,  the  given 

condition  is 

PC  =  r. 

Third  step.    By  (lY),  p.  31, 


PC  =  V(cc  -  ay  +(y-  pf. 
.-.  ^{x  -  af  +  {y-  ^y  =  r. 
Squaring  and  transposing,  we  have  (II).  q.e.d. 

Corollary.  The  equation  of  the  circle  whose  center  is  the  origin 
(0,  0)  and  whose  radius  is  r  is 

a?2  +  i/2  =  J.2, 

The  following  facts  should  be  observed : 

Any  straight  line  is  defined  by  an  equation  of  the  first  degree 
in  the  variables  x  and  y. 

Any  circle  is  defined  by  an  equation  of  the  second  degree  in 
the  variables  x  and  y,  in  which  the  terms  of  the  second  degree 
consist  of  the  sum  of  the  squares  of  x  and  y. 


THE  CURVE  AND  THE  EQUATION         69 

30.  Locus  of  an  equation.  The  preceding  sections  have  illus- 
trated the  fact  that  a  locus  problem  in  Analytic  Geometry  leads 
p,t  once  to  an  equation  in  the  variables  x  and  y.  This  equation 
having  been  found  or  being  given,  the  complete  solution  of  the 
locus  problem  requires  two  things,  as  already  noted  in  the  first 
section  (p.  51)  of  this  chapter,  namely, 

1.  To  draw  the  locus  by  plotting  a  sufficient  number  of  points 
whose  coordinates  satisfy  the  given  equation,  and  through  which 
the  locus  therefore  passes. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve. 

These  two  problems  are  respectively  called : 

1.  Plotting  the  locus  of  an  equation  (second  fundamental 
problem). 

2.  Discussing  an  equation  (third  fundamental  problem). 

For  the  present,  then,  we  concentrate  our  attention  upon  some 
given  equation  in  the  variables  x  and  y  (one  or  both)  and  start 
out  with  the  definition  : 

The  locus  of  an  equation  in  two  variables  repTCsenting  coordinates 
is  the  curve  or  group  of  curves  passing  through  all  points  whose 
coordinates  satisfy  that  equation,"*  and  through  such  points  only. 

From  this  definition  the  truth  of  the  following  theorem  is  at 
once  apparent : 

Theorem  III.  If  the  form  of  the  given  equation  he  changed  in  any 
way  (^for  example,  by  transposition,  by  multiplicatidn  by  a  constant, 
etc.),  the  locus  is  entirely  unaffected. 

*  An  equation  in  the  variables  x  and  y  is  not  necessarily  satisfied  by  the  coordinates  of 
any  points.  For  coordinates  are  real  numbers,  and  the  form  of  the  equation  may  be  such 
that  it  is  satisfied  by  no  real  values  of  x  and  y.    For  example,  the  equation 

is  of  this  sort,  since,  when  x  and  ij  are  real  numbers,  x^  and  y^  are  necessarily  positive 
(or  zero),  and  consequently  x^  +  y^  +  lis  always  a  positive  number  greater  than  or  equal 
to  1,  and  therefore  not  equal  to  zero.  Such  an  equation  therefore  has  no  locus.  The 
expression  "the  locus  of  the  equation  is  imaginary"  is  also  used. 

An  equation  may  be  satisfied  by  the  coordinates  of  a  finite  number  of  points  only. 
For  example,  x^  +  y'^=0  is  satisfied  by  x=0,  y  =  0,  but  by  no  other  real  values.  In  this 
case  the  group  of  points,  one  or  more,  whose  coordinates  satisfy  the  equation,  is  called 
the  locus  of  the  equation. 


36^ 


60 


ANALYTIC  GEOMETRY 


We  now  take  up  in  order  the  solution  of  the  second  and  third 
fundamental  problems. 

31.  Second  fundamental  problem. 

Rule  to  plot  the  locus  of  a  given  equation. 

First  step.  Solve  the  given  equation  for  one  of  the  variables  in 
terms  of  the  other. ^ 

Second  step.  By  this  formula  compute  the  values  of  the  vari- 
able for  which  the  equation  has  been  solved  by  assuming  real 
values  for  the  other  variable. 

Third  step.  Plot  the  points  corresponding  to  the  values  so 
determined.^ 

Fourth  step.  If  the  points  are  numerous  enough  to  suggest  the 
general  shape  of  the  locus,  draw  a  smooth  curve  through  the  poi^its. 

Since  there  is  no  limit  to  the  number  of  points  which  may  be 
computed  in  this  way,  it  is  evident  that  the  locus  may  be  drawn 
as  accurately  as  may  be  desired  by  simply  plotting  a  sufficiently 
large  number  of  points. 

Several  examples  will  now  be  worked  out  and  the  arrangement 
of  the  work  should  be  carefully  noted. 

Ex.  1.    Draw  the  locus  of  the  equation 

2x-Sy  +  6  =  0. 
Solution.    First  step.    Solving  for  y, 

y  =  |x  +  2. 
Second  step.    Assume  values  for  x  and  compute 
y,  arranging  results  in  the  form  : 


YJi 

ji 

^ 

Y 

y 

K 

y 

r 

[y 

(0, 

2; 

y 

^ 

> 

y 

(--3 

0) 

0 

X 

2\ 

Thus,  if 


1,  y  =  2  .  1  +  2 

2,  y  .r  1 .  2  +  2 

etc. 


31, 


Third  step.    Plot  the  points  found. 
Fourth   step.     Draw    a   smooth    curve 
through  these  points. 


X 

y 

X 

y 

0 

2 

0 

2 

1 

2| 

-  1 

n 

2 

3^ 

-2 

f 

3 

4 

-3 

0 

4 

42- 

-4 

_    2 
.3 

etc. 

etc. 

etc. 

etc. 

*  The  form  of  the  given  equation  will  often  be  such  that  solving  for  one  variable  is 
simpler  than  solving  for  the  other.    Always  choose  the  simpler  solution. 
t  Remember  that  real  values  only  may  be  used  as  coordinates. 


THE  CURVE  AND  THE  EQUATION 


61 


Ex.  2.    Plot  the  locus  of  the  equation 

2/  zz  x2  —  2  x  —  o. 

Solution.    First  step.    The  equation  as  given  is  solved  for  y. 
Second  step.    Computing  y  by  assuming  values  of  x,  we  find  the  table  of 
v^alues  below : 


X 

y 

X 

y 

0 

-3 

0 

-3 

.1 

-4 

-  1 

0 

2 

—  o 

-2 

5 

3 

0 

-3 

12 

4 

5 

-4 

21 

5 

12 

etc. 

etc. 

6 

21 

etc. 

etc. 

i^ 

I 

/ 

/ 

/ 

f 

/ 

V 

/ 

\ 

/ 

\ 

/ 

f 

y 

L 

0 

i 

V 

\ 

/ 

/ 

s^ 

/ 

' 

I 

Third  step.    Plot  the  points. 

Fourth  step.    Draw  a  smooth  curve  through  these  points.     This  gives  the 
curve  of  the  figure. 

Ex.  3.    Plot  the  locus  of  the  equation 

x2  +  ^2  4_  e  X  -  IG  =  0. 
First  step.    Solving  for  y, 

y  =  ±  VlG  -  6  X  -  x2. 
Second  step.    Compute  y  by  assuming  values  of  x. 


X 

y 

X 

y 

0 

±4 

0 

±4 

1 

±3 

,  - 1 

±4.6 

2 

0 

—  2 

±4.9 

3 

imag. 

-3 

±5 

4 

" 

-4 

±4.9 

5 

u 

-5 

±4.6 

G 

u 

-6 

±4 

7 

u 

-7 

±3 

-8 

0 

-9 

imag. 

I  Fa     I      I 

X'        "y" o      J~^ 


62  ANALYTIC  GEOMETRY 

For  example,  if  x  =  1,  y  =  ±  Vl6  -6-1  =  ±  3  ; 

if  X  =  3,  y  =  ±  Vl6  -  18  -  9  =  ±  V- 11, 
an  imaginary  number ; 

if  X  =  -  1,  y  =  ±  Vl6  +  6-1  =  ±  4.6, 
etc. 

Third  step.    Plot  the  corresponding  points. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 

PROBLEMS 

1.  Plot  the  locus  of  each  of  the  following  equations, 
(a)  X  +  2  ?/  =  0. 
(b)x  +  22/  =  3. 

(c)  3  X  -  ?/  +  5  =  0. 

(d)  2/  =  4  x2. 

(e)  x2  +  4  y  =  0. 

(f)  2/  =  x2  -  3. 

(g)  x2  +  4  y  -  5  =  0. 
(h)  2/  =  x2  +  X  +  1. 
(i)  a;  =  2/2  +  22/-3. 
(j)  4x  =  y3. 
(k)  4  X  =  2/3-1. 

(1)  y  =  x^-  1. 
(m)  y  =  x^  —  X. 
(n)  ?/  =  x^  —  x2  —  5. 


(P)    X2  +  y2  :=  9. 

(q)  x2  +  2/2  =  25. 

(r)  x2  +  2/2  +  9a.. 

=  0. 

(s)  x2  +  2/2  +  4  2/  = 

=  0. 

(t)  x2  +  2/2  -  6  X  - 

-16: 

=  0. 

(u)  x2  +  2/2  -  6  2/  - 

-16: 

=  0. 

(v)  4  2/  =  x4  -  8. 

(w)  4  X  =  2/*  +  8. 

/^\  .,  _      ^ 

^''^^-1+X2 

"i"^:- 

(z)x  =  -?-. 

r 


(o)  x2  +  2/2  =  4.  1  +  2/ 

2.  Show  that  the  following  equations  have  no  locus  (footnote,  p.  59). 

(a)  x2  +  2/2  +  1  =  0.  (f )  x2  +  2/2  +  2  X  +  2  2/  +  3  =  0. 

(b)  2x2  +  32/2^-8.  (g)  4x2  4-2/2  +  8x  +  5  =  0. 

(c)  x2  +  4  =  0.  (h)  ?/*  +  2  x2  +  4  =  0. 

(d)  x* +  2/2  + 8  =  0.  (i)  9x2+4y24-i8x+82/  +  15=0. 

(e)  (X  +  1)2  +  2/2  +  4  =  0.  (i)  x2  +  X2/  +  2/2  +  3  =  0. 

Hint.    Write  each  equation  in  the  form  of  a  sum  of  squares,  or  solve  for  one  variable 
and  apply  Theorem  III,  p.  11,  to  the  quadratic  under  the  radical. 

32.  Principle  of  comparison.     In  Ex.  1,  p.  60,  and  Ex.  3,  p.  61, 

we  can  determine  the  nature  of  the  locus,  that  is,  discuss  the 
equation,  by  making  use  of  the  formulas  (I)  and  (II),  p.  58.  The 
method  is  important  and  is  known  as  the  principle  of  comparison. 


THE  CURVE  AND  THE  EQUATION        63 

The  nature  of  the  locus  of  a  given  equation  may  he  determined 
by  comparison  with  a  general  known  equation,  if  the  latter  becomes 
■  identical  with  the  given  equation  by  assigning  particular  values  to 
its  coefficients. 

The  method  of  making  the  comparison  is  explained  in  the 
following 

Rule.  First  step.  Change  the  form*  of  the  given  equation  (if 
necessary)  so  that  one  or  more  of  its  terms  shall  be  identical  with 
one  or  more  terms  of  the  general  equation. 

Second  step.  Equate  coefficients  of  corresponding  terms  in  the 
two  equations,  supplying  any  terms  missing  in  the  given  equation 
with  zero  coefficients. 

Third  step.  Solve  the  equations  found  in  the  second  step  for 
the  values  f  of  the  coefficients  of  the  general  equation. 

Ex.  1.  Show  that  2x  — 3y  +  6  =  0  is  the  equation  of  a  straight  line 
(Fig.,  p.  60).     " 

Solution.    First  step.    Compare  with  the  general  equation  (I),  p.  68, 

(1)  y  =  mx  +  b. 

Put  the  given  equation  in  the  form  of  (1)  by  solving  for  y, 

(2)  y=fx  +  2. 

Second  step.  The  right-hand  members  are  now  identical.  Equating 
cCefl&cients  of  x, 

(3)  m  =  f . 
Equating  constant  terms, 

(4)  6  =  2. 

Third  step.  Equations  (3)  and  (4)  give  the  values  of  the  coefficients  m 
and  6,  and  these  are  possible  values,  since,  p.  34,  the  slope  of  a  line  may- 
have  any  real  value  whatever,  and  of  course  the  ordinate  b  of  the  point 
(0,  b)  in  which  a  line  crosses  the  F-axis  may  also  be  any  real  number.  There- 
fore the  equation  2x-3y  +  6  =  0  represents  a  straight  line  passing  through 
(0,  2)  and  having  a  slope  equal  to  §.  q.e.d. 

*This  transformation  is  called  "putting  the  given  equation  in  the  form"  of  the 
general  equation. 

tThe  values  thus  found  may  be  impossible  (for  example,  imaginary)  values.  This 
may  indicate  one  of  two  things,  —  that  the  given  equation  has  no  locus,  or  that  it  cannot 
be  put  in  the  form  required. 


64  ANALYTIC  GEOMETRY 

Ex.  2.    Show  that  the  locus  of 

(5)  x2  +  2/2 +  Gx- 16  =  0 
is  a  circle  (Fig.,  p.  61). 

Solution.    First  step.    Compare  with  the  general  equation  (II),  p.  68, 

(6)  x2  +  2/2  -  2  ax  -  2  jSy  +  a2  +  /32  -  r2  =  0. 

The  right-hand  members  of  (5)  and  (6)  agree,  and  also  the  first  two  terms, 
x2  +  ?/2. 

Second  step.    Equating  coefficients  of  x, 

(7)  -  2  a  =  6. 
Equating  coefficients  of  y, 

(8)  -2/3  =  0. 
Equating  constant  terms, 

(9)  a2  + /32  -  r2  ==  -  16.  ^ 
Third  step.    From  (7)  and  (8), 

a  =  -  3,  /3  =  0. 

Substituting  these  values  in  (9)  and  solving  for  r,  we  find 
r2  =  25,  or  r  =  5. 

Since  a,  /3,  r  may  be  any  real  numbers  whatever,  the  locus  of  (5)  is  a 
circle  whose  center  is  (—  3,  0)  and  whose  radius  equals  5. 


PROBLEMS 

1.  Plot  the  locus  of  each  of  the  following  equations.  Prove  that  the  locus 
is  a  straight  line  in  each  case,  and  find  the  slope  m  and  the  point  of  inter- 
section with  the  axis  of  y,  (0,  6). 


(a) 

2x4-2/-6  = 

0. 

(b) 

X-Sy  +  S  = 

0. 

(c) 

x  +  2y  =  0. 

(d) 

5  X  -  6  ?/  -  5 

=  0. 

(e) 

i^-fy-i 

=  0. 

(f) 

?-f-l=0 
5      6 

is) 

7  X  -  8  ?y  =  0. 

(h) 

f^-fy-l 

=  0. 

Ans.  m  =—  2,  b  =  6. 

Ans.  m  =  1,  6  =  2|. 

Ans.  m  =—  I,  b  =  0. 

Ans.  m.  =  I,  6  =  —  |. 


Ans. 

m  =  l,b=-j\. 

Ans. 

m  =  |,  6  =  -6. 

Ans. 

m=  I,  6  =  0. 

Ans. 

rn=l,b  =  -l^. 

THE  CURVE  AND  THE  EQUATION        65 

2.  Plot  the  locus  of  each  of  the  equations  following,  and  prove  that  the 
locus  is  a  circle,  finding  the  center  (or,  /3)  and  the  radius  r  in  each  case.^ 

(a)  x2  +  2/2  _  16  =  0.  Ans.  {a,  p)  =  (0,  0);  r  =  4. 

(b)  x2  +  y2  _  49  =  0.  Ans.  (or,  ^)  =  (0,  0);  r  =  7. 

(c)  aj2  +  ?/2  -  25  =  0.  Ans.  {a,  p)  =  {0,0);  r  =  5. 

(d)  x2  +  ?/2  +  4x  =  0.  Ans.  (or,  /3)  =  (-  2,  0);  r  =  2. 

(e)  x2  +  2/'^  -  8  7/  =  0.  Ans.  {a,  p)  =  (0,  4);  r  =  4.      _ 

(f)  x2  +  2/2  +  4x-8?/  =  0.  Ans.  {a,  p)  =  {- 2,4);  r  =  ^20. 

(g)  x2  +  2/2  _  6x  4-  4 y  -  12  =  0.  Ans.  (a,  ^3)  =  (3,  -  2);  r  =  5. 
(h)  x2  +  2/2  -  4x  +  92/  -  I  =  0.  Ans.  (a,  ^3)  =  (2,  -  |);  r  =  5. 

(i)  3x2  +  32/2-6x- 82/  =  0.  ^ns.   (a,  ^)  =  (1,|);  r  =  f. 

The  following  problems  illustrate  cases  in  which  the  locus 
problem  is  completely  solved  by  analytic  methods,  since  the  loci 
may  be  easily  drawn  and  their  nature  determined. 

3.  Find  the  equation  of  the  locus  of  a  point  whose  distances  from  the 
axes  XX'  and  YY'  are  in  a  constant  ratio  equal  to  f . 

A  ns.    The  straight  line  2  x  —  3  y  =  0. 

4.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  whose  distances 
from  the  axes  of  coordinates  is  always  equal  to  10. 

Ans.   The  straight  line  x  +  y  —  10  =  0. 

5.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  (3,  0)  and  (0,  —  2)  is  always  equal  to  8.  Find  the  equation  of  the 
locus  and  plot. 

Ans.    The  parallel  straight  lines  6x  +  42/  +  3  =  0,  6x  +  42/-13  =  0. 

6.  A  point  moves  so  as  to  he  always  equidistant  from  the  axes  of  coor- 
dinates.    Find  the  equation  of  the  locus  and  plot. 

Ans.    The  perpendicular  straight  lines  x-\-y  =  0,x  — y  =  0. 

7.  A  point  moves  so  as  to  be  always  equidistant  from  the  straight  lines 
jc  -  4  =  0  and  2/  +  5  =  0.     Find  the  equation  of  the  locus  and  plot. 

Ans. .  The  perpendicular  straight  lines  x-y-9  =  0,  x  +  y-\-l=0. 

8.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  the  squares  of 
whose  distances  from  (3,  0)  and  (-3,  0)  always  equals  68.     Plot  the  locus. 

Ans.    The  circle  x2  +  2/2  =  25. 

9.  Find  the  equation  of  the  locus  of  a  point  which  moves  so  that  its  dis- 
tances from  (8,  0)  and  (2,  0)  are  always  in  a  constant  ratio  equal  to  2.  Plot 
the  locus.  Ans.    The  circle  x2  +  2/^  =  16. 

10.  A  point  moves  so  that  the  ratio  of  its  distances  from  (2, 1)  and  (-  4,  2) 
is  always  equal  to  |.     Find  the  equation  of  the  locus  and  plot. 

Ans.    The  circle  3  x2  +  3  2/2  -  24  x  -  4  2/  =  0. 


i.:> 


66  ANALYTIC  GEOMETRY 

In  the  proofs  of  the  following  theorems  the  choice  of  the  axes 
of  CQordinates  is  left  to  the  student,  since  no  mention  is  made 
of  either  coordinates  or  equations  in  the  problem.  In  such  cases 
always  choose  the  axes  in  the  most  convenient  manner  possible. 

11.  A  point  moves  so  that  the  sum  of  its  distances  from  two  perpendicular 
lines  is  constant.     Show  that  the  locus  is  a  straight  line. 

Hint.  Choosing  the  axes  of  coordinates  to  coincide  with  the  given  lines,  the  equation 
ia  x  +  y=  constant. 

12.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  two  fixed  points  is  constant.     Show  that  the  locus  is  a  straight  line. 

Hint.  Draw  XX^  through  the  fixed  points,  and  Y  Y''  through  their  middle  point.  Then 
the  fixed  points  may  be  written  (a,  0),  (-  a,  0),  and  if  the  "constant  difference  "  be  denoted 
by  k,  we  find  for  the  locus  4  ax  =  ^'  or  4  ax  =  -  fc. 

13.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
two  fixed  points  is  constant.     Prove  that  the  locus  is  a  circle. 

Hint.   Choose  axes  as  in  problem  12. 

14.  A  point  moves  so  that  the  ratio  of  its  distances  from  two  fixed  points 
is  constant.     Determine  the  nature  of  the  locus. 

Ans.  A  circle  if  the  constant  ratio  is  not  equal  to  unity  and  a  straight 
line  if  it  is. 

The  following  problems  illustrate  the 

Theorem.  If  an  equation  can  be  put  in  the  form  of  a  product  of 
variable  factors  equal  to  zero,  the  locus  is  found  by  setting  each  fac- 
tor equal  to  zero  and  plotting  the  locus  of  each  equation  separately. 

16.  Draw  the  locus  of        4  x2  -  9  2/2  :=  0. 
Solution.    Factoring, 

(1)  {2x-Sy){2x  +  3y)  =  0. 

Then,  by  the  theorem,  the  locus  consists  of  the  straight  lines 

(2)  2x-3y  =  0, 

(3)  2x-\-Sy  =  0. 

Proof.  1.  The  coordinates  of  any  point  (xi,  ?/i)  which  satisfy  (1)  will 
satisfy  either  (2)  or  (3). 

For  if  (xi,  yi)  satisfies  (1), 

(4)  (2xi-32/i)(2xi  +  32/i)  =  0. 


THE  CURVE  AND  THE  EQUATION         67 

This  product  can  vanish  only  when  one  of  the  factors  is  zero.     Hence 

either 

2a;i- 32/1  =  0, 

and  therefore  (xi,  yi)  satisfies  (2) ; 

or  2  xi  +  3  yi  =  0, 

and  therefore  (xi,  yi)  satisfies  (3). 

2.  A  point  (xi,  yi)  on  either  of  the  lines  defined  by  (2)  and  (3)  will  also 
lie  on  the  locus  of  (1). 

For  if  (xi,  yi)  is  on  the  line  2  x  -  3  y  =  0, 
then  (Corollary,  p.  53) 
(5)  2xi-3?/i  =  0. 

Hence  the  product  (2  xi  —  3  yi)  (2  xi  +  3  yi)  also  vanishes,  since  by  (5)  the 
first  factor  is  zero,  and  therefore  (xi,  yi)  satisfies  (1). 

Therefore  every  point  on  the  locus  of  (1)  is  also  on  the  locus  of  (2)  and 
(3),  and  conversely.     This  proves  the  theorem  for  this  example.  q.e.d. 

16.  Show  that  the  locus  of  each  of  the  following  equations  is  a  pair  of 
straight  lines,  and  plot  the  lines. 

(a)  z^-y^=:  0.  (j)  3x2  +  xy  -  22/2  +  6x  -  4y  =  0. 

(b)  9X2  -2/2  =  0.  (k)   x2  -  2/2  +  X  +  ?/  =  0. 

(c)  x2  =  92/2.  (1)  x2  -  X2/  +  5x  -  5 2/  =  0. 

(d)  x2  -  4  X  -  5  =  0.  (m)  x2  -  2  X2/  +  2/2  +  6  X  -  6  2/  =  0. 

(e)  2/2-6?/  =  7.  (n)  x2 -42/2+ 5x+ 102/  =  0. 

(f)  2/2  -  5xy  +  62/  =  0.  (o)  x2  +  4x2/  +  4  2/2  +  5x  +  IO2/  +  6  =  0. 

(g)  X2/  -  2x2  -  3x  =  0.  .(p)  x2  +  3x2/  +  2 2/2  +  X  +  2/  =  0. 

(h)  X2/  -  2  X  =  0.  (q)  x2  -  4  X2/  -  5  2/2  +  2  X  -  10  2/  =  0. 

(i)  X2/  =  0.  (r)  3  x2  -  2  X2/  -  2/2  +  5  X  -  5  y  =  0. 

17.  Show  that  the  locus  of  Ax"^  +  Bx  +  C  =  0  is  a  pair  of  parallel  lines,  a 
single  line,  or  that  there  is  no  locus  according  as  A  =  B^  —  4:  AC  is  positive, 
zero,  or  negative. 

18.  Show  that  the  locus  of  ^x2  +  Bxy  +  Cy^  =  0  is  a  pair  of  intersecting 
lines,  a  single  line,  or  a  point  according  as  A  =  B^  —  4:  AC  is  positive,  zero, 

or  negative.  ' '  ' '  ^  '•  •  -a^w^  ,  ^/ 

\     33.  Third  fundamental  problem.    Discussion  of  an  equation. 

The  method  explained  of  solving  the  second  fundamental  prob- 
lem gives  no  knowledge  of  the  required  curve  except  that  it 
passes  through  all  the  points  whose  coordinates  are  determined 
as  satisfying  the  given  equation.  Joining  these  points  gives  a 
curve  more  or  less  like  the  exact  locus.     Serious  errors  may  be 


68 


ANALYTIC  GEOMETRY 


made  in  this  way,  however,  since  the  nature  of  the  curve  between 
any  two  successive  points  plotted  is  not  determined.  This  obj  ection 
is  somewhat  obviated  by  determining  before  plotting  certain  prop- 
erties of  the  locus  by  a  discussion  of  the  given  equation  now  to 
be  explained. 

The  nature  and  properties  of  a  locus  depend  upon  the  form  of 
its  equation,  and  hence  the  steps  of  any  discussion  must  depend 
upon  the  particular  problem.  In  every  case,  however,  the  fol- 
lowing questions  should  be  answered. 

1.  Is  the  curve  a  closed  curve  or  does  it  extend  out  infinitely  far? 

2.  Is  the  curve  symmetrical  with  respect  to  either  axis  or  the 
origin  ? 

The  method  of  deciding  these  questions  is  illustrated  in  the 
following  examples. 


Ex.  1.    Plot  the  locus  of 


IP 


(1) 


x2  +  4  2/2  =  16. 


Discuss  the  equation. 
Solution.    First  step.    Solving  for  x, 
(2)  x  =  ±2  V4  -  ?/2. 

Second  step.    Assume  values  of  y  and  compute  x.    This  gives  the  table. 

Third  step.    Plot  the  points  of  the  table. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 


X 

y 

X 

y 

±4 

0 

±4 

0 

±3.4 

1 

±3.4 

-1 

±2.7 

H 

±2.7 

-H 

0 

2 

0 

-2 

imag. 

3 

imag. 

-3 

Discussion.  1.  Equation  (1)  shows  that  neither  x  nor  y  can  be  indefi- 
nitely great,  since  x^  and  4  ?/2  are  positive  for  all  real  values  and  their  sum 
must  equal  16.  Therefore  neither  x^  nor  4?/2  can  exceed  16.  Hence  the 
curve  is  a  closed  curve. 

A  second  way  of  proving  this  is  the  following : 

From  (2),  the  ordinate  y  cannot  exceed  2  nor  be  less  than  —  2,  since  the 
expression  4  —  ^/2  beneath  the  radical  must  not  be  negative.  (2)  also  shows 
that  X  has  values  only  from  —  4  to  4  inclusive. 


yu  ;^ 


THE  CURVE  AND  THE  EQUATION 


69 


2.  To  determine  the  symmetry  with  respect  to  the  axes  we  proceed  as 
follows : 

The  equation  (1)  contains  no  odd  powers  of  x  or  y ;  hence  it  may  be  writ- 
ten in  any  one  of  the  forms 

(3)  (x)2  4-  4  (-  2/)2  =  16,  replacing  (x,  y)  by  (x,  -  y) ; 

(4)  (-  x)2  +  4  (?/)2  =  16,  replacing  (x,  y)  by  (-  x,  y) ; 

(5)  (-  x)2  +  4  (-  y)2  =  16,  replacing  (x,  ?/)  by  (-  x,  -  y). 

The  transformation  of  (1)  into  (3)  corresponds  in  the  figure  to  replacing 
each  point  P(x,  y)  on  the  curve  by  the  point  Q(x,  —  y).  But  the  points  P 
and  Q  are  symmetrical  with  respect  to  XX\  and  (1)  and  (3)  have  the  same 
locus  (Theorem  HI,  p.  59).  Hence  the  locus  of  (1)  is  unchanged  if  each  point 
is  changed  to  a  second  point  symmetrical  to  the  first  with  respect  to  XX\ 
Therefore  the  locus  is  symmetrical  with  respect  to  the  axis  of  x.  Similarly 
from  (4),  the  locus  is  symmetrical  with  respect  to  the  axis  ofy,  and  from  (5), 
the  locus  is  symmetrical  with  respect  to  the  origin. 

The  locus  is  called  an  eUipse. 

Ex.  2.    Plot  the  locus  of 

(6)  y2_4a;_f.i5  =  o. 

Discuss  the  equation. 

Solution.  First  step.  Solve  the  equation  for  x,  since  a  square  root  would 
have  to  be  extracted  if  we  solved  for  y.     This  gives 

(7)  X  =  i(y2  +  15). 


X 

y 

33- 

0 

4 

±1 

^ 

±2 

6 

±3 

U 

±4 

10 

±5 

12| 

±6 

etc. 

etc. 

Second  step.    Assume  values  for  y  and  compute  x. 


70 


ANALYTIC  GEOMETRY 


Since  y"^  only  appears  in  the  equation,  positive  and  negative  values  of  y 
give  the  same  value  of  x.     The  calculation  gives  the  table  on  p.  69. 

For  example,  if  2/  =  ±  3, 

then  a;  =  i  (9  +  15)  =  6,  etc. 

Third  step.    Plot  the  points  of  the  table. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 

Discussion.    1.  From  (7)  it  is  evident  that  x  increases  as  y  increases. 

Hence  the  curve  extends  out  indefinitely  far  from  both  axes. 

2.  Since  (6)  contains  no  odd  powers  of  y,  the  equation  may  be  written  in 

the  form  -        \o       ^  /  \   ,  nr      r. 

(-?/)2-4(x)  +  15  =  0 

by  replacing  (x,  y)  by  (x,  —  y).     Hence  the  locus  is  symmetrical  with  respect 
to  the  axis  of  x. 

The  curve  is  called  a  parabola. 

Ex.  3.    Plot  the  locus  of  the  equation 
(8)  xy-2y-4  =  0.  '{? 

Solution.    First  step.    Solving  for  y, 

4 

(9) 


x-2 


Second  step.    Compute  ?/,  assuming  values  for  x. 

When 


X 

y 

X 

y 

0 

-2 

0 

-2 

1 

-4 

-1 

-f 

n 

-8 

-2 

-1 

If 

-16 

-4 

-f 

2 

cp 

-5 

-4 

2i 

16 

: 

2| 

8 

-10 

s 

3 

4 

etc. 

.  etc. 

4 

2 

5 

4 

6 

1 

12 

0.4 

etc. 

etc. 

2,  2/  =  1  =  00. 

In  such  cases  we  assume  values  differing 
slightly  from  2,  both  less  and  greater,  as  in 
the  table. 

Third  step.    Plot  the  points. 

Fourth  step.  Draw  the  curve  as  in  the 
figure  in  this  case,  the  curve  having  two 
branches. 

1.  From  (9)  it  appears  that  y  diminishes 
and  approaches  zero  as  x  increases  indefi- 
nitely. The  curve  therefore  extends  indefi- 
nitely far  to  the  right  and  left,  approaching 
constantly  the  axis  of  x.  If  we  solve  (8)  for 
X  and  write  the  result  in  the  form 
4 


X  =  2 


y 


it  is  evident  that  x  approaches  2  as  ?/  increases 
indefinitely.     Hence  the  locus  extends  both 
upward  and  downward  indefinitely  far,  approaching  in  each  case  the  line  x  =2. 


THE  CURVE  AND  THE  EQUATION 


71 


v/ 


2.  The  equation  cannot  be  transformed  by  any  one  of  the  three  substitutions 
(X,  y)  into  (x,  -  y), 
(X,  y)  into  (-x,  y), 
(X,  y)  into  (-  x,  -  ?/), 

without  altering  it  in  such  a  way  that  the  new  equation  will  not  have  the 
same  locus.  The  locus  is  therefore  not  symmetrical  with  respect  to  either 
axis,  nor  with  respect  to  the  origin. 


— 

— 

— 

— 

— 

— 

oo 

— 

— 

— 

— 

— 

— 

— 

— 

^ 

\ 

v^ 

^^ 

^ 

'^ 

p 

oo 

^ 

=— 



u 

0 

0 

,0) 

5: 

*> 

N 

1 

-00 

_ 

_ 

_ 

_ 

_ 

_ 

_ 





This  curve  is  called  an  hyperbola. 

Ex.  4.    Draw  the  locus  of  the  equation 
(10)  4y  =  x^ 


X 

y 

X 

y 

0 

0 

0 

0 

1 

i 

-1 

-\ 

H 

-II 

-H 

-'n 

2 

2 

-2    ' 

-2 

n 

^% 

-2i 

-3|| 

3 

H 

-3 

-6| 

3i 

lOfl 

-3f 

-lOff 

Solution.    First  step.    Solving  for  y, 
y  =  1x3. 

Second  step.  Assume  values  for  x 
and  compute  y.  Values  of  x  must  be 
taken  between  the  integers  in  order  to 
give  points  not  too  far  apart. 

For  example,  if 

x  =  2|, 

2/  =  |.i|5=J^2_5=3i|,  etc. 


72 


ANALYTIC  GEOMETRY 


Kk 

i 

1 

1 

X 

? 

; 

/ 

/ 

V 

' 

,/^ 

/O 

V 

// 

// 

r\ 

f/ 

' 

f-x,- 

J 

- 

- 

--I 

— 

- 

Third  step.    Plot  the  points  thus  found. 
Fourtli  step.    The  points  determine  the  curve  of  tlie 
figure.       •; 

Discussion.  1.  From  the  given  equation  (10),  x  and 
y  increase  simultaneously,  and  therefore  the  curve 
extends  out  indefinitely  from  both  axes. 

2.  In  (10)  there  are  no  even  powers  nor  constant 
term,  so  that  by  changing  signs  the  equation  may  be 
written  in  the  form 

4(-2/)  =  (-x)3, 

replacing  (x,  ?/)  by  {-x,  -  y). 

Hence  the  locus  is  symmetrical  with  respect  to  the 
origin. 

The  locus  is  called  a  cubical  parabola. 


34.  Symmetry.  In  the  above  examples  we  have  assumed  the 
definition : 

If  the  points  of  a  curve  can  be  arranged  in  pairs  which  are 
symmetrical  with  respect  to  an  axis  or  a  point,  then  the  curve 
itself  is  said  to  be  symmetrical  with  respect  to  that  axis  or  point. 

The  method  used  for  testing  an  equation  for  symmetry  of  the 
locus  was  as  follows  :  if  (x,  y)  can  be  replaced  by  {x,  —  y)  through- 
out the  equation  without  affecting  the  locus,  then  if  (a,  b)  is  on 
the  locus,  (a,  —  h)  is  also  on  the  locus,  and  the  points  of  the  latter 
occur  in  pairs  symmetrical  with  respect  to  XX\  etc.     Hence 

Theorem  IV.  If  the  locus  of  an  equation  is  unaffected  by  replacing 
y  hy  —  y  throughout  its  equation,  the  locus  is  symmetrical  with 
respect  to  the  axis  of  x. 

If  the  locus  is  unaffected  by  changing  x  to  —  x  throughout  its 
equation,  the  locus  is  symmetrical  with  respect  to  the  axis  of  y. 

If  the  locus  is  unaffected  by  changing  both  x  and  y  to  —  x  and 
—  y  throughout  its  equation,  the  locus  is  symmetrical  with  respect 
to  the  origin. 

These  theorems  may  be  made  to  assume  a  somewhat  different 
form  if  the  equation  is  algebraic  in  x  and  y  (p.  17).  The  locus 
of  an  algebraic  equation  in  the  variables  x  and  y  is  called  an 
algebraic  curve.     Then  from  Theorem  lY  follows 


THE  CURVE  AND  THE  EQUATION        73 

Theorem  V.  Symmetry  of  an  algebraic  curve.  If  no  odd  powers 
of  y  occur  in  an  equation^  the  locus  is  symmetrical  with  respect  to 
XX';  if  no  odd  powers  of  x  occur,  the  locus  is  symmetrical  with 
respect  to  YY'.  If  every  term  is  of  even*  degree,  or  every  term  of 
odd  degree,  the  locus  is  symmetrical  with  respect  to  the  origin. 

35.  Further  discussion.  In  this  section  we  treat  of  three,  more 
questions  which  enter  into  the  discussion  of  an  equation. 

3.  Is  the  origin  on  the  curve  ? 
This  question  is  settled  by 

Theorem  VI.  The  locus  of  an  algebraic  equation  passes  through 
the  origin  when  there  is  no  constant  term  in  the  equation. 

Proof  The  coordinates  (0,  0)  satisfy  the  equation  when  there 
is  no  constant  term.  Hence  the  origin  lies  on  the  curve  (Corol- 
lary, p.  53).  Q.E.D. 

4.  What  values  of  x  and  y  are  to  be  excluded  ? 
Since  coordinates  are  real  numbers  we  have  the 

Rule  to  determine  all  values  of  x  and  y  which  must  be  excluded. 

First  step.  Solve  the  equation  for  x  in  terms  of  y,  and  from  this 
result  determine  all  values  of  y  for  which  the  computed  value  of  x 
will  be  imaginary.     These  values  of  y  must  be  excluded. 

Second  step.  Solve  the  equation  for  y  in  terms  of  x,  and  from 
this  result  determine  all  values  of  x  for  which  the  computed  value 
of  y  will  be  imaginary.     These  values  of  x  must  be  excluded. 

The  intercepts  of  a  curve  on  the  axis  of  x  are  the  abscissas  of 
the  points  of  intersection  of  the  curve  and  XX\ 

The  intercepts  of  a  curve  on  the  axis  of  y  are  the  ordinates  of 
the  points  of  intersection  of  the  curve  and  YT. 

Rule  to  find  the  intercepts. 

Substitute  y  —  0  and  solve  for  real  values  of  x.  This  gives  the 
intercepts  on  the  axis  of  x. 

Substitute  x  =  0  and  solve  for  real  values  of  y.  This  gives  the 
intercepts  on  the  axis  of  y. 

*  The  constant  term  must  be  regarded  as  of  even  (zero)  degree. 


74 


ANALYTIC  GEOMETRY 


The  proof  of  the  rule  follows  at  once  from  the  definitions. 
The  rule  just  given  explains  how  to  answer  the  question: 
5.  What  are  the  intercepts  of  the  locus  ? 

36.  Directions  for  discussing  an  equation.  Given  an  equation, 
the  following  questions  should  be  answered  in  order  before  plot- 
ting the  locus. 

1.  Is  the  origin  on  the  locus?  (Theorem  VI). 

2.  Is  the  locus  sijmmetrical  ivith  respect  to  the  axes  or  the 
origin?  {^Theorems  IV and  V). 

3.  What  are  the  intercepts?  (Mule,  p.  73). 

4.  What  values  of  x  and  y  must  he  excluded  ?  (^Rule,  p.  73). 

5.  Is  the  curve  closed  or  does  it  pass  off  indefinitely  far?  (§  33, 
p.  68). 

Answering  these  questions  constitutes  what  is  called  a  general 
discussion  of  the  given  equation. 

Ex.  1.    Give  a  general  discussion  of  the  equation 
(1)  x2  -  4  2/2  +  16  2/  =  0. 

Draw  the  locus. 


X 

Sr 

n 

k 

^ 

* 

^ 

^^ 

X 

■Vs 

^ 

^ 

^ 

^ 

Vw 

_^ 

^ 

^ 

— ' 

^ 

^ 

!l= 

i 

fo, 

0 

y= 

2 

X 

^ 

^" 

'o 

^ 

X 

^ 

\^ 

*^ 

k^ 

^ 

L^ 

^ 

^ 

. 

* 

■^ 

* 

r' 

X 

1.  Since  the  equation  contains  no  constant  terra,  the  origin  is  on  the  curve. 

2.  The  equation  contains  no  odd  powers  of  x\  hence  the  locus  is  symmet- 
rical with  respect  to  YY'. 

3.  Putting  ?/  =:  0,  we  find  cc  =  0,  the  intercept  on  the  axis  of  x.     Putting" 
ic  =  0,  we  find  ?/  =  0  and  4,  the  intercepts  on  the  axis  of  y. 

4.  Solving  for  x,  

(2)  x.=  ±2V2/2_4y. 


THE  CURVE  AND  THE  EQUATION        75 

Hence  all  values  of  y  between  0  and  4  must  be  excluded,  since  for  such  a 
value  y2  _  4y  is  negative  (Theorem  IH,  p.  11). 

Solving  for  y, 
(3)  2/  =  2  ±  i  Vx--2  +  16. 

Hence  no  value  of  x  is  excluded,  since  x^  +  16  is  always  positive. 

5.  From  (3),  y  increases  as  x  increases,  and  the  curve  extends  out 
indefinitely  far  from  both  axes. 

Plotting  the  locus,  using  (2),  the  curve  is  found  to  be  as  in  the  figure. 
The  curve  is  an  hyperbola. 

PROBLEMS 

1.  Give  a  general  discussion  of  each  of  the  following  equations  and  draw 
the  locus. 

(n)  9?/2_x3  =  0. 
\o)  9?/2  +  x3  =  0. 
(p)  2xy +  3x-4  =:0. 
(q)  x2  -  X2/  +  8  =  0. 
(r)  x2  +  xy  -4  =  0. 
(s)  x2  +  2x?/-3y  =  0.' 
^>(t)  2x?/-?/3  +  4x  =  0. 
'^  (u)  3x2-?/  +  x  =  0. 
(v)  4  2/2  -  2  X  -  y  =  0. 
(w)  x2  -  2/2  +  6  X  =  0. 
(x)  x2  +  4  2/2  +  8  ?/  =  0. 
^y)  9x2  +  2/2  +  I8x  -  Qy  =  0. 
>(z)  9x2-2/2  +  18x  +  6?/  =  0. 

2.  Determine  the  general  nature  of  the  locus  in  each  of  the  following 
equations  by  assuming  particular  values  for  the  arbitrary  constants,  but  not 
special  values,  that  is,  values  which  give  the  equation  an  added  peculiarity.* 

(a)  2/2  =  2  mx.  (f )  x^  -  y^  =  a^. 

(b)  x2  -  2  my  =  m2.  (g)  x^  +  y^  =  r^. 
x2      2/2  _  (h)  x2  +  2/2  =  2rx. 

(°)  ^+52-^-  (i)  x2  +  2/2  =  2r2/. 

(d)  2x2/  =  «2.  (J)  ^'  +  y^  =  2ax  +  2by. 

3,2      ^2  (k)  a2/2  =  x^ 
(^)  a^-b2  =  ^-  (1)  a%  =  a:3. 

*  For  example,  in  (a)  and  (b)  m=  0  is  a  special  value.  In  fact,  in  all  these  examples 
zero  is  a  special  value  for  any  constant. 


(a)  x2  - 

-42/  =  0. 

(b)  2/2  - 

-4x  +  3  =  0. 

(c)  x2  + 42/2 -16  =  0. 

(d)  9x2 

+  2/''  -  18  =  0. 

(e)  x2  - 

-  4  2/2  -  16  =  0. 

(f )  x2  - 

-42/2  +  16  =  0. 

(g)  x^  - 

-2/2  +  4  =  0. 

(h)    X2  - 

-  2/  +  X  =  0. 

(i)  xy  - 

-4  =  0. 

(j)  9  2/  +  ic3  =  0. 

(k)  4x 

-  2/3  =  0. 

.  (1)  6x 

-y'=^o. 

(m)  5x 

-  2/  +  2/=^  =  0. 

76  ANALYTIC  GEOMETRY 

3.  Draw  the  locus  of  the  equation 

y'^  =  [x  —  a)  (x  —  6)  (x  -  c), 

(a)  when  a  <  6  <  c.  (c)  when  a  <h,h  =  c. 

(b)  when  a  =  h  <c.  (d)  when  a=^h  =  c. 

The  loci  of  the  equations  (a)  to  (f )  in  problem  2  are  all  of  the 
class  known  as  conies,  or  conic  sections,  —  curves  following  straight 
lines  and  circles  in  the  matter  of  their  simplicity. 

A  conic  section  is  the  locus  of  a  point  whose  distances  from  a 
fixed  point  and  a  fixed  line  are  in  a  constant  ratio. 

4.  Show  that  every  conic  is  represented  by  an  equation  of  the  second 
degree  in  x  and  y. 

Hint.  Take  YY'  to  coincide  with  the  fixed  line,  and  draw  XX'  through  the  fixed  point. 
Denote  the  fixed  point  by  {p,  0)  and  the  constant  ratio  by  e. 

Ans.   (1  -  e2)x2  +  ?/2  -  2j)x  +  p2  -  o. 

5.  Discuss  and  plot  the  locus  of  the  equation  of  problem  4, 

(a)  when  e  =  \.     The  conic  is  now  called  a  parabola  (see  p.  70). 

(b)  when  e  <  1.     The  conic  is  now  called  an  ellipse  (seep.  69). 

(c)  when  e  >  1.     The  conic  is  now  called  an  hyperbola  (see  p.  71). 

6.  Plot  each  of  the  following. 

(i)  x  = 

(J)  ^  = 


(a)  x2?/  -  5  =  0. 

(e)  y 

(b)  x2y  -  ?/  +  2  X  =  0. 

(f)  y 

(c)  x?/2_4x4-6  =  0. 

(g)  y 

(d)  T'y  -y  +  ^  =  0. 

(h)  y 

x2-3x 
4x2 

x2-4 
x-3 

x  +  1 
x2-4 

2/-1 

y-2 


(k)4x  =  -/ 
8y 


X2+  X 


(1)    X 


37.  Points  of  intersection.  If  two  curves  whose  equations 
are  given  intersect,  the  coordinates  of  each  point  of  intersection 
must  satisfy  both  equations  when  substituted  in  them  for  the 
variables  (Corollary,  p.  53).  In  Algebra  it  is  shown  that  all 
values  satisfying  two  equations  in  two  unknowns  may  be  found 
by  regarding  these  equations  as  simultaneous  in  the  unknowns' 
and  solving.     Hence  the 

Rule  to  find  the  points  of  intersection  of  two  curves  ivhose  equa- 
tions are  given. 


THE  CURVE  AND  THE  EQUATION 


77 


First  step.  Consider  the  equations  as  simultaneous  in  the  coordi- 
nates, and  solve  as  in  Algebra. 

Second  step.  Arrange  the  real  solutions  in  corresponding  pairs. 
These  will  he  the  coordinates  of  all  the  points  of  intersection. 

Notice  that  only  real  solutions  correspond  to  common  points 
of  the  two  curves,  since  coordinates  are  always  real  numbers. 

Ex.  1.    Find  the  points  of  intersection  of 

(1)  a: -7?/ +  25  =  0, 

(2)  x2  +  2/2  =  25. 

Solution.    First  step.    Solving 
(1)  for  X, 

(3)  x.=  1y-2b. 
Substituting  in  (2), 

{ly-  25)2  +  2/2  =  25. 
Reducing,   2/^  —  7  ?/  +  12  =  0. 

.-.  2/  =  3  and  4. 
Substituting  in  (3)  [not  in  (2)], 

X  =  —  4  and  +  3. 

Second  step.  Arranging,  the  points  of  intersection  are  (—4,  3)  and 
(3,  4).     Ans. 

In  tlie  figure  the  straight  line  (1)  is  the  locus  of  equation  (1),  and  the 
circle  the  locus  of  (2). 

Ex.  2.    Find  the  points  of  intersection  of  the  loci  of 


(4) 

(5) 


2x2  +  32/2  =  35, 
3x2-42/  =  0. 


Solution.    First  step.    Solving  (5)  for  x2, 
(6)  x2  =  f2/. 

Substituting  in  (4)  and  reducing, 
92/2 +  82/ -105  =  0. 

.-.  y  =  S  and  —  %^ 

Substituting  in  (6)  and  solving, 
X  =  ±  2  and  ±  |  V-  105. 

Second  step.  Arranging  the  real  values,  we  find  the  points  of  intersection 
are  (+  2,  3),  (-  2,  3).     Ans. 

In  the  figure  the  ellipse  (4)  is  the  locus  of  (4),  and  the  parabola  (5)  the 
locus  of  (5). 


/ 

I 

it 

(■■JsX^  — 

-J^ 

^'^y_ 

7% 

t    \ 

z  ^ 

-I      \ 

^      \-* 

x^         ^ 

it^ 

s; 

7 

^)s 

f 

r' 

"^ 

78  ANALYTIC   GEOMETRY 

PROBLEMS 

Find  the  points  of  intersection  of  the  following  loci. 
,     7x-ll2/  +  l  =  01 

""■   x-y  =  ^}'  Ans.i6,l). 

^-   l^Vy^tl}'  ^'''-  (0'2),  (-1,  -I). 

^     2/2  ^  16  X  ^ 

4-     y-.x  =  Of'  ^^'-    (^'^)'    (l^'l^)- 

x2  +  2/2  _  4 X  +  6 y  -  12  =  0^ 
^'   2y  =  Sx  +  3  }•  ^^MTV,m(-3,  -3). 

^       x2-?/2  =  16^ 

^'    x2  =  8y         }•  Ans.   i±4V2,i). 

^     x^  +  y^  =  41'] 

^-    xy  =  20         r  ^^^-  (±5'  ±4),  (±4,  ±6). 

x2  +  2/2_6a;_2y_i5  =  o       1  ,        ,     ^    ..    , 

^-    9x2  +  92/2  +  6x-62/-27  =  0r  ^^5-   ("  2,  1),  (- H,  -  ff). 

10.  ^ .    For  what  values  of  6  are  the  curves  tangent  ? 


/-«.  ( ^ , ),  6  =  ±7VTa 

y2  --  2  rix  ^ 
^^-    x^  =  2pyf'  -  ^'^'-   (0,  0),  (2j),  2p). 

,-     4x2  +  ?/2^5>l 
12-    y2  33  8x  r  ^'''-   (-2),  (1,-2). 

x2  z=  4  ay        1  - 

^^-   y^      ^^^      r  ^ns.  (2a,  a),  (-2a,  a). 

x2  +  4a2j 

x2  +  ?/2^100^ 

!*•    ^2-^  h  ^ws.   (8,6),  (8,  -6). 

2  J 

,_     x^  +  y^  =  5a^-] 

•    x2  =  4ay         r  ^''^-   (2a,  «),  (-2a,  a). 

,„       6%2+  (j2y2  =  ^262>| 

^®'    X2  +  2/2  =  (^2  |-  ^^s-   («,  0),  (-  a,  0). 


THE  CURVE  AND  THE  EQUATION         79 

17.  The  two  loci  — -^  =  1  and  ?-  +  ^  =  4  intersect  in  four  points. 

4        9  4        9 

Find  the  lengths  of  the  sides  and  of  the  diagonals  of  the  quadrilateral  formed 
by  these  points. 

Ans.   Points,  (±  VlO,  ±  |  V6).    Sides,  2  VlO,  3  Vc.    Diagonals,  V94. 

Find  the  area  of  the  triangles  and  polygons  whose  sides  are  the  loci  of  the 
following  equations. 

18.  3x  +  y +  4  =  0,  3x-5?/  +  34  =  0,  3a;-2?/  +  l  =  0.  Ans.  36. 

19.  x-\-2y  =  6,2x  +  y  =  1,y  =  x-\-l.  '          Ans.  f. 

20.  x  +  y  =  a,x-2y  =  ia,  y-x-\-1a  =  0.  Ans.  12 a^. 

21.  x  =  0,y  =  0,x  =  4,y  =-6.  Ans.  24. 

22.  x-y  =  0,  x-\-y  =  0,  x  —  y  =  a,  x  +  y  =  b.  Ans.   — . 

2 

23.  2/  =  3x-9,  ?/  =  3x  + 5,  2y  =x-6,  2y  =  a;  +  14.        Ans.    56. 

24.  Find  the  distance  between  the  points  of  intersection  of  the  curves 
3x  -  2 ?/  +  6  =  0,  x2  +  2/2  =  9.  Ans.    if  ^^• 

25.  Does  the  locus  of  y^  =  4x  intersect  the  locus  of  2x  +  3y  +  2  =  0? 

Ans.    Yes. 

26.  For  what  value  of  a  will  the  three  lines  3x  +  ?/  —  2  =  0,  ax +  2?/  —  3  =  0, 
2x  —  2/  —  3  =  0  meet  in  a  point  ?  Ans.    a  =  5. 

27.  Find  the  length  of  the  common  chord  of  x^  -\-  y^  =  lii  and  ?/2  =  3  x  +  3. 

Ans.    6. 

28.  If  the  equations  of  the  sides  of  a  triangle  are  x  +  Ty  +  11  =  0, 
3x  +  2/— 7  =  0,  X  —  3?/  +  l  =  0,  find  the  length  of  each  of  the  medians. 

Ans.    2  V5,  I  V2,  1  VlTO. 

Show  that  the  following  loci  intersect  in  two  coincident  points,  that  is,  are 
tangent  to  each  other. 

29.  ?/2-10x-6?/-31  =0,  2?/-10x  =  47. 

30.  9x2-4?/2  +  54x-162/  +  29  =  0,  15x-8?/  +  ll  =  0. 

38.  Transcendental  curves.  The  equations  thus  far  consid- 
ered have  been  algebraic  in  x  and  ?/,  since  powers  alone  of  the 
variables  have  appeared.  We  shall  now  see  how  to  plot  certain 
so-called  transcendental  curves,  in  which  the  variables  appear 
otherwise  than  in  powers.     The  Rule,  p.  60,  will  be  followed. 


80 


ANALYTIC  GEOMETKY 


Ex.  1.    Draw  the  locus  of 

(1)  y  =  \ogiox. 

Solution.    Assuming  values  for  x,  y  may  be  computed  by  a  table  of  loga- 
rithms, or,  remembering  the  definition  of  a  logarithm,  from  (1)  will  follow 

(2)  X  =  lO^'. 

Hence  values  may  also  be  assumed  for  y,  and  x  computed  by  (2).     This 
is  done  in  the  table. 
In  plotting, 

unit  length  on  XJT'  is  2  divisions, 
unit  length  on  YY'  is  4  divisions. 

General  discussion.  1.  The  curve  does  not 
pass  through  the  origin,  since  (0,  0)  does  not 
satisfy  the  equation. 

2.  The  curve  is  not  symmetrical  with  re- 
spect to  either  axis  or  the  origin. 
3.  In  (1),  putting  x  =  0, 

2/  =  log  0  =  —  00  =  intercept  on  YY\ 
In  (2),  putting  y  =  0, 

X  =  100  =  I  =  intercept  on  XX'. 
Ya 


X 

y 

X 

y 

1 

0 

.1 

-  1 

3.1 

i 

.01 

-2 

10 

1 

.001 

-3 

100 

2 

.0001 

-4 

etc. 

etc. 

etc. 

etc. 

4.  From  (2),  since  logarithms  of  negative  numbers  do  not  exist,  all  nega- 
tive values  of  x  ofre  excluded. 

From  (2)  no  value  of  y  is  excluded. 

5.  From  (2),  as  y  increases  x  increases,  and  the  locus  extends  out  indefi- 
nitely from  both  axes. 

From  (1),  as 

X  approaches  zero, 

y  approaches  negative  infinity ; 
so  we  see  that  the  curve  extends  down  indefinitely  and  approaches  nearer 
and  nearer  to  YY\ 


THE  CURVE  AND  THE  EQUATION 


81 


Ex.  2.    Draw  the  locus  of 
(3)  y  =  8mx 

if  the  abscissa  x  is  the  circular  measure  of  an  angle  (Chapter  I,  p.  19). 

Solution.    Assuming  values  for  x  and  finding  the  corresponding  number 
of  degrees,  we  may  compute  y  by  the  table  of  Natural  Sines,  p.  21. 
For  example,  if 

X  =  1,  since  1  radian  =  57°.  29, 

y  =  sin  57°. 29  =  .843.  [by  (3)] 

It  will  be  more  convenient  for  plotting  to  choose  for  x  such  values  that 
the  corresponding  number  of  degrees  is  a  whole  number.  Hence  x  is 
expressed  in  terms  of  ;r  i«i  the  table. 

For  example,  if 


X 

y 

X 

y 

0 

0 

0 

0 

7t 

6 

.50 

It 
~  6 

-.50 

7t 

3 

.86 

It 
~  3 

-.86 

It 
2 

1.00 

It 

~  2 

-1.00 

27r 
"3" 

.80 

27r 
3 

-.80 

57r 
6 

.50 

hit 
6 

-.50 

Tt 

0 

-  It 

0 

,       y  =  sni  - 

2Tt 

— ,  ?/  =  sm 


sin  60° 


2Tt 


-sin  ?^  (4,  p.  19) 


=  -  sin  120°=  -sin 60°  (5, p.  20) 
=  -  .86. 
In  plotting,  three  divisions  being  taken 
as  the  unit  of  length,  lay  off 

•  ^0  =  OB  =  7r  =  3.1416, 
and  divide  AO  and  OB  up  into  six  equal 
parts. 

The  course  of  the  curve  beyond  B  is 
easily  determined  from  the  relation 
sin  (2  ;r  +  x)  =  sin  x. 
Hence     y  =  smx  =  sin (2  ;r  +  x), 
that  is,  the  curve  is  unchanged  ifx  +  27tbe  substituted  for  x.     This  means, 
however,  that  every  point  is  moved  a  distance  2  tt  to  the  right.    Hence  the  arc 


r. 

__ 

f^ 

,j 

V 

y 

■■1 

# 

^ 

V 

^ 

*■ 

> 

S; 

s. 

, 

-J 

TT- 

f' 

f 

0 

/ 

1 

1 

e. 

't" 

'3, 

/ 

' 

1 

1 

^^ 

y 

X 

1 

^ 

^ 

4\J 

'■ 

\ 

/ 

TT 

TT 

IT 

Itt 

5jr 

n* 

A^ 

\ 

" 

" 

SL 

1 

1 

1 

^ 

-- 

£. 

i- 

.i>. 

-- 

a._ 

.T 

p-: 

— 

*s 

- 

- 

- 

— 

f/ 

^ 

^ 

^ 

s 

•^ 

^ 

~ 

~ 

~ 

p 

?/= 

-1 

/? 

~^ 

r 

V' 

- 

APO  may  be  moved  parallel  to  XX'  until  A  falls  on  B,  that  is,  into  the 
position  BEC,  and  it  will  also  be  a  part  of  the  curve  in  its  new  position. 


82 


ANALYTIC  GEOMETRY 


Also,  the  arc  OQB  may  be  displaced  parallel  to  XX'  until  0  falls  upon  C.  In 
this  way  it  is  seen  that  the  entire  locus  consists  of  an  indefinite  number  of 
congruent  arcs,  alternately  above  and  below  XX'. 

General  discussion.    1.  The  curve  passes  through  the  origin,  since  (0,  0) 
^satisfies  the  equation. 

2.  Since  sin (—  x)  =  —  sin x,  changing  signs  in  (3), 


or 


y  =—  sm  X, 
y  =  sin(-x). 


Hence  the  locus  is  unchanged  if  (x,  y)  is  replaced  by  (— x,  —y),  and 
the  curve  is  symmetrical  with  respect  to  the  origin  (Theorem  IV,  p.  72) . 

3.  In  (3),  if  X  =  0, 

y  =  sin  0  =  0  =  intercept  on  the  axis  of  y. 
Solving  (3)  for  x, 
(4)  x  =  sin-'^y. 

In  (4),  if  y  =  0, 


=  me,  n  being  any  integer. 

Hence  the  curve  cuts  the  axis  of  x  an  indefinite  number  of  times  both  on 
the  right  and  left  of  0,  these  points  being  at  a  distance  of  tt  from  one  another. 
4.  In  (3),  X  may  have  any  value,  since  any  number  is  the  circular  meas- 
ure of  an  angle. 

In  (4),  y  may  have  values  from  -  1  to  +  1  inclusive,  since  the  sine  of  an 
angle  has  values  only  from  —  1  to  +1  inclusive. 

5.  The  cuTve  extends 
out  indefinitely  along  XX' 
in  both  directions,  but  is 
contained  entirely  between 
the  lines  ?/=:  +  l,  y  =  —i. 
The  locus  is  called  the 
wave  curve,  from  its  shape, 
or  the  sinusoid,  from  its 
equation  (3). 

Ex.  3.  Draw  the  locus 
of  y  =  tan  x. 

There  is  no  difficulty  in 
obtaining  the  curve  of  the' 
figure  and  in  verifying  the 
properties  indicated  by  a  dis- 
cussion similar  to  the  pre- 
ceding examples. 


THE  CURVE  AND  THE  EQUATION  83 

PROBLEMS 

Plot  the  loci  of  the  following  equations. 

1.  2/  =  cosx.  5.    y  =  ta,n-^x.  9.    y  =  sin2x. 

2.  y  =  cotx.  6.   2/ =  2^.  10.   2/  =  tan-. 

3     y-s&cx  '^-    y  =  21ogioX. 

3.  y-secx.  ^  11     y  =  2co8X. 

4.  y  =  sm-^x.  8.    2/  =  (l+ajf.  12.   y  =  sin x  +  cos x. 

39.  Graphical  representation  in  general.  Any  equation  con- 
taining two  variables  may  be  represented  graphically  by  a  curve 
called  the  graph  of  the  equation  by  considering  the  variables  as 
coordinates  and  plotting  the  locus  in  the  usual  way.  This 
method  of  representing  a  given  law  is  widely  used  in  all  branches 
of  science. 

Ex.  1.    Draw  the  graph  of  the  Simple  Interest  Law,  which  shall  represent 
the  relation  between  amount  and  time  for  a  given  principal  and  rate  per  cent. 
The  law  is  proven  in  Algebra  to  be 

(1)  J.  =  P(l  +  m), 

where     A  =  amount,  P  =  principal,  r  =  rate,  n  =  number  of  years. 

Solution.    For  convenience,  take  P  =  one  dollar.*    Let 
One  division  on  OX  =  1  year. 
One  division  on  OF  =  1  dollar, 

abscissas  =  values  of  n, 

ordinates  =  values  of  A. 

(0,1) 

Then  the  required  graph  is  the  locus  of 

O  (1,0)     (n,o)X 

(2)  ?/  =  rx  +  1. 

The  locus  of  (2)  is  a  straight  line  passing  through  (0,  1)  and  having  a 
slope  equal  to  r  (Theorem  I,  p.  58). 

This  graph  may  be  used  to  solve  interest  problems.  For  if  the  number  of 
years  n  is  given,  we  merely  have  to  measure  off  the  corresponding  ordinate 
A  of  the  straight  line,  and  this  will  give  the  amount  of  one  dollar  at  the  given 
rate  for  n  years. 

*  Any  other  case  is  obtained  by  multiplying  all  the  ordinates  in  the  figure  by  P. 


84  ANALYTIC  GEOMETRY 

Ex.  2.  In  Physics  it  is  shown  that  the  vokime  (v),  pressure  (p),  and  absolute 
temperature  (t)  of  a  given  mass  of  a  perfect  gas  are  connected  by  the  law 

(3)  pv  =  kt, 

k  being  a  constant  dependent  upon  the  particular  gas. 
Draw  the  graph  if  the  temperature  is  assumed  constant. 

Solution.    Assume 

one  division  on  OX  =  unit  of  pressure, 
one  division  on  OY  =  unit  of  volume, 
abscissas  =  pressures, 
ordinates  =  volumes. 

Then  the  required  graph  is  the  locus  o^ 

(4)  xy  =  constant. 

The  curve  is  one  branch*  of  an  hyperbola  extending  to  the  right  and 
upward  indefinitely,  approaching  in  each  case  the  corresponding  axis.  Such 
curves  are  called  isothermals  (equal  temperatures),  and  jtl^e  figure  is  called 
the  Pressure-Volume  Diagram. 

PROBLEMS 

1.  Draw  the  graph  of  the  Simple  Interest  Law  if  the  variables  are 

(a)  n  and  P.  (c)  A  and  P.  (e)  P  and  r. 

(b)  n  and  r.  (d)  A  and  r. 

2.  Draw  the  graph  of  the  law  of  Ex.  2  if  the  variables  are 

(a)  p  and  t.  (b)  v  and  t. 

3.  The  amount  (A)  of  any  principal  (P)  at  compound  interest  (r%)  for  n 
years  is  given  by  the  Compound  Interest  Law 

A=P{1  +  r)». 
Draw  the  graph  of  this  law  if  the  variables  are 

(a)  A  and  P.  (c)  A  and  n.  (e)  P  and  n. 

(b)  A  and  r.  ■  (d)  P  and  r.  (f)  r  and  n. 

Hint.    Take  tlie  logarithm  of  both  sides  when  convenient  for  computation. 

*  Since  negative  volumes  have  no  physical  meaning,  in  many  cases  only  a  portion  of 
the  entire  locus  can  be  made  use  of  in  the  representation. 


CHAPTER  IV 

THE  STRAIGHT  LINE  AND  THE   GENERAL  EQUATION  OF 
THE   FIRST  DEGREE 

40.  The  idea  of  coordinates  and  the  intimate  relation  connect- 
ing a  curve  and  an  equation,  which  results  from  the  introduction 
of  coordinates  into  the  study  of  Geometry,  have  been  considered 
in  the  preceding  chapters.  Analytic  Geometry  has  to  do  largely 
with  a  more  detailed  study  of  particular  curves  and  equations. 
In  this  chapter  we  shall  consider  in  detail  the  straight  line  and 
the  general  equation  of  the  first  degree  in  the  variables  x  and  y 
representing  coordinates. 

41.  The  degree  of  the  equation  of   a  straight  line.   It  was 

shown  in  Chapter  III  (Theorem  I,  p.  58)  that 

(1)  y  =  mx  +  b 

is  the  equation  of  the  straight  line  whose  slope  is  m  and  whose 
intercept  on  the  F-axis  is  ^ ;  m  and  b  may  have  any  values, 
positive,  negative,  or  zero  (p.  34).  But  if  a  line  is  parallel  to 
the  F-axis,  its  equation  may  not  be  put  in  the  form  (1) ; ,  for, 
in  the  first  place,  the  line  has  no  intercept  on  the  F-axis,  and, 
in  the  second  place,  its  slope  is  infinite  and  hence  cannot  be 
substituted  for  m  in  (1).  The  equation  of  a  line  parallel  to  the 
F-axis  is,  however,  of  the  form 

(2)  X  =  constant. 

The  equation  of  any  line  may  be  put  either  in  the  form  (1)  or 
(2).  As  these  equations  are  both  of  the  first  degree  in  x  and  y 
we  have 

Theorem  I.  The  expiation  of  any  straight  line  is  of  the  first  degree 
in  the  coordinates  x  and  y. 

85 


86  ANALYTIC  GEOMETRY 

42.  The  general  equation  of  the  first  degree,  A3o+By-\-C=0, 

The  equation 

(1)  Ax-{-By-\-C  =  0, 

where  A,  B,  and  C  are  arbitrary  constants  (p.  1),  is  called  the 
general  equation  of  the  first  degree  in  x  and  y  because  every  equa- 
tion of  the  first  degree  may  be  reduced  to  that  form. 
Equation  (1)  represents  all  straight  lines. 

For  the  equation  y  =  mz  +  b  may  be  written  mx  —  y  -{■  b  -  0,  which  is  of  the 
form  (1)  it  A  =  m,  B  =  —  1,  C=  b;  and  the  equation  x  =  constant  may  be  written 
X  —  constant  =  0,  which  is  of  the  form  (1)  if  J.  =  1,  ^  =  0,  C=  —  constant. 

Theorem  II.  (Converse  of  Theorem  I.)  The  locus  of  the  general 
equation  of  the  first  degree 

Ax -{-  By  -{- C  =^  0 
is  a  straight  line.  X? 

Proof    Solving  (1)  for  y,  we  obtain 

AC 

(2)  2/  =  -^--^- 

This  equation  has  the  same  locus  as  (1)  (Theorem  111,  p.  59). 
By  Theorem  I,  p.  58,  the  locus  of  (2)  is  the  straight  line  whose 

A  C 

slope  is  m  =  —  —  and  whose  intercept  on  the  F-axis  is  ft  =  —  —  • 
B  B 

If,  however,  £  =  0,  it  is  impossible  to  write  (1)  in  the  form 

(2).     But  if  ^  =  0,  (1)  becomes 

^x  +  C  =  0, 

C 

or  x= 7  • 

A 

The  locus  of  this  equation  is  a  straight  line  parallel  to  the 
F-axis  (1,  p.  57).  Hence  in  all  cases  the  locus  of  (1)  is  a  straight 
line.  Q.E.D. 

Corollary  I.    The  slope  of  the  line 

Ax -[-  By -{- C  =  () 

is  m  =—  —;    that  is,  the  coefficient  of  x  with  its  sign  changed 
B 

divided  by  the  coefficient  of  y. 


1^  THE  STRAIGHT  LINE  87 

Corollary  II.    The  lines 

Ax -{-  Bi/  -\- C  =  0 
and  A^x-\- B'y  +  C  =  0 

are  parallel  when  and  only  when  the  coefficients  of  x  and  y  are 
proportional;  that  is, 

A__B 
A'^B'' 

For  two  lines  are  parallel  when  and  only  when  their  slopes  are  equal  (Theorem 
VI,  p.  36) ;  that  is,  when  and  only  when 

_A__A^ 
B~      B'' 
Changing  the  signs  and  applying  alternation,  we  obtain 

Corollary  III.    The  lines 

Ax-^By  -\-C  =  0 

and  A'x -\- B'y  +  C  =^  0 

are  perpendicular  when  and  only  when 

AA'  +  BB'  =  0. 

For  two  lines  are  perpendicular  when  and  only  when  the  slope  of  one  is  the 
negative  reciprocal  of  the  slope  of  the  second  (Theorem  VI,  p.  36) ;  that  is, 

A  _B' 
~B-A^' 
or  AA'  +  BB'  =  0. 


Ly  - 


Corollary  IV.    The  intercepts  of  the  line 
Ax  +  By  -^C  =  0 
on  the  X-  and  Y-axes  are  respectively 

a  = 7  and  0  =  — -  • 

A  B 

For  the  intercept  on  the  X-axis  is  found  (p.  73)  by  setting  y  =  (i  and  solving 
for  X,  and  the  intercept  on  the  F-axis  has  been  found  in  the  above  proof. 

Corollaries  I  and  IV  are  given  chietly  for  purposes  of  reference.  In  a  numerical 
example  the  intercepts  are  found  most  simply  by  applying  the  general  rule  already 
given  (p.  73) ;  and  the  slope  is  found  by  reducing  the  equation  to  the  form 

2/  =  mx  +  &, 
when  the  coefficient  of  x  will  be  the  slope. 


88  ANALYTIC  GEOMETRY 

Theorems  I  and  II  may  be  stated  together  as  follows : 
The  locus  of  an  equation  is  a  straight  line  when  and  only  when 
the  equation  is  of  the  first  degree  in  x  and  y. 

Theorem  II  asserts  that  the  locus  of  every  equation  of  the  first 
degree  is  a  straight  line.  Then,  to  plot  the  locus  of  an  equation 
of  the  first  degree  it  is  merely  necessary  to  plot  two  points  on  the 
locus  and  draw  the  straight  line  passing  through  them.  The  two 
simplest  points  to  plot  are  those  at  which  the  line  crosses  the 
axes.  But  if  those  points  are  very  near  the  origin  it  is  better  to 
use  but  one  of  them  and  some  other  point  not  near  the  origin 
whose  coordinates  are  found  by  the  Eule  on  p.  60. 

Theorem  III.     When  two  equations  of  the  first  degree, 

(3)  Ax  +  By  ^  C  =  0  ; 
and 

(4)  A^x'+B'y  +  C^  =  0, 

have  the  same  locus,  then  the  corresjjonding  coefficients  are  propor- 
tional; that  is, 

A'~~  B'~  C' 

Proof  The  lines  whose  equations  are  (3)  and  (4)  are  by 
hypothesis  identical  and  hence  they  have  the  same  slope  and  the 
same  intercept  on  the  F-axis.     Since  they  have  the  same  slope, 

A       A' 

B^B''  (Corollary  I,  p.  86) 

and  since  they  have  the  same  intercept  on  the  F-axis, 

B~  B' 
by  alternation  we  obtain 


=  — .  (Corollary  lY,  p.  87) 


A        B        ^    C       B 
A'  =  J'''''^C'  =  B''' 

,  ,  ABC 

and  hence  -  =  -=-.  q.^..^. 


THE  STRAIGHT  LINE,  89 

Ex.  1.    Find  the  values  of  a  and  h  for  which  the  equations 
2ax  +  2?/-5  =  0 
and  4x-3?/  +  76  =  0 

will  represent  the  same  straight  line. 

Solution.  These  two  equations  will  represent  the  same  straight  line  if 
(Theorem  III)  2a  _  ^  _  -5  . 

4    ~  -3~  76  ' 
and  hence  the  required  values  are  obtained  by  solving 

2a_^  _2___  -b^ 

4    ~  33  ^"     33  ~  "75" 
for  a  and  h.     This  gives 

a  =  -t,  6  =  ||. 

43.  Geometric  interpretation  of  the  solution  of  two  equations 
of  the  first  degree.    If  we  solve  the  equations 

(1)  Ax  +  By  -\-C  =  0 
and 

(2)  A^x -{- B^y -\- C^  =  0, 

we  obtain  the  coordinates  of  the  points  of  intersection  of  the 
lines  whose  equations  are  (1)  and  (2)  (Eule,  p.  76).  But  if 
these  lines  are  parallel  they  do  not  intersect,  and  if  they  are 
identical  they  intersect  in  all  of  their  points.  The  relation 
between  the  position  of  the  lines  whose  equations  are  (1)  and 
(2)  and  the  number  of  solutions  of  the  simultaneous  equations 
(1)  and  (2)  may  be  indicated  as  follows : 

_,    .  .        ^  _  Number  of  solutions 

Fosition  of  Lines 

of  equations 

Intersecting  lines.  One  solution. 

Parallel  lines.  No  solution. 

Coincident  lines.  An  infinite  number. 

It  is  sometimes  as  convenient  to  be  able  to  determine  the 
number  of  solutions  of  two  equations  of  the  first  degree  without 
solving  them  as  it  is  to  be  able  to  determine  the  nature  of  the 
roots  of  a  quadratic  equation  without  solving  it.  The  following 
theorem  enables  us  to  do  this. 


90  ANALYTIC  GEOMETRY 

Theorem  IV.    Two  equations  of  the  first  degree^ 
Ax-\-  By  +  C  =  0 
and  A'x -{- B'y -\- C^  =  0, 

have,  in  general,  one  solution  for  x  and  y ;  hut  if 

A__B_ 
A'~  B'' 
there  is  no  solution  unless 

A__B__C_ 
A'~B'~C'' 

when  there  is  an  infinite  number  of  solutions. 

The  proof  follows  at  once  from  Corollary  II,  p.  87,  and  Theorem  III. 

PROBLEMS  V 

1 .  Find  the  intercepts  of  the  following  lines  and  plot  the  lines. 

(a)  2  x  +  3  2/  =  6.  Ans.  3,  2. 

(b)  ^  +  ^  =  1.  '  Ans.  2,  4. 
2      4 

(c)  -  -  ^  =  1.  Ans.  3,  -  5. 

(d)  -  +  -^  =  1.  Ans.  4,  -2. 
4       —  2 

2.  Plot  the  following  lines. 

(a)  2x-32/+5  =  0.  (c)  ^  +  |  =  1. 

(b)  2/-5-4ic  =  0.  (d)  --y-  =  \. 

3.  Find  the  equations,  and  reduce  them  to  the  general  form,  of  the  lines 
for  which 

(a)  m  =  2,  6  =  -  3.  ^ns.  2  x  -  y  -  3  =  0. 

(b)  m  =  -  i,  6  =  f .  Ans.  x  +  2  ?/  -  3  =  0. 

(c)  m  =  f ,  6  =  -  f .  Ans.  4  x  -  10  y  -  25  =  0. 

(d)  a  =  - ,  6  =  —  2.  Ans.  x-y  -2  =  0. 

4 

q  _, 

(e)  or  =  — - ,  6  =  3.  Ans.  x  +  y  -  3  =  0. 
Hint.    Substitute  \ny  =  mx  +  b. 


THE  STRAIGHT  LINE  91 

A.  Find  the  number  of  solutions  of  the  following  pairs  of  equations  and 
plot  the  loci  of  the  equations. 

(^)  I4  X  +  6  2/  +  9  =  0.  ^'''-  ^°  '^^^*^°°- 


(b)  { 


/  Ans.  One. 

x  +  y  =  1. 

'2-3x 


(d)  -^  .   ^      ^  Ans.  No  solution. 

I^lz  X  —  lo ' 


('2 3x^1/ 

(c)  -^  -      ,0        ■  ^ns.  An  infinite  number. 

^  '    1^6  X  +  2  ?/  =  4. 

20  =  0. 

2/4-6  =  0. 

5.  Plot  the  lines  2x  —  32/  +  6  =  0  and  x  —  y  =  0.     Also  plot  the  locus  of 
(2  X  -  3  2/  +  6)  +  A;  (X  -  y)  =  0  for  A:  =  0,  ±1,  ±.2.     , 

6.  Select  pairs  of  parallel  and  perpendicular  lines  from  the  following. 
fLi:y  =  2x-S. 

,,JX2:?/=-3X+2.  .  TUT  T  ,       T 

^^nU:y  =  2x  +  7.  ^^'-  Li\\Ls;L,±L,. 

L2>4  :  y  =  1 X  +  4. 
rXi:x  +  3y  =  0. 

(b)  ^  X2  :  8x  +  y  +  1  =  0.  Ans.  Li  ±  L3. 

[Ls-.dx-Sy  +  2  =  0. 
fXi :  2  X  -  5  y  =  8. 

(c)  -^  X2 :  5  y  +  2  X  =  8.  Ans.  X2  -L  L3. 

U3:35x-14y  =  8. 

7.  Show   that    the    quadrilateral    whose    sides   are    2x  — 3y  +  4  =  0, 
3x  —  y  —  2  =  0,  4x  —  6?/  —  9  =  0,  and  6x  —  22/  +  4  =  0isa  parallelogram. 

8.  Find  the  equation  of  the  line  whose  slope  is  —  2  which  passes  through 
the  point  of  intersection  of  y  =  3  x  +  4  and  y  =  —  x  +  4. 

Ans.  2x  +  2/  —  4  =  0. 

9.  What  is  the  locus  of  y  =  wx  +  6  if  6  is  constant  and  m  arbitrary  ?  if 
m  is  constant  and  b  arbitrary  ? 

10.  Write  an  equation  which  will  represent  all  lines  parallel  to  the  line 
(a)  y  =  2  X  +  7.  (c)  y  -  3  X  -  4  =  0. 

(b)?/  =  -x  +  9.  (d)  22/-4x  +  3  =  0. 

11.  Write  an  equation  which  will  represent  all  lines  having  the  same 
intercept  on  the  F-axis  as  (a),  (b),  (c),  and  (d)  in  problem  10. 

12.  Find  the  equation  of  the  line  parallel  to2x  —  3?/  =  0  whose  intercept 
on  the  F-axis  is  —  2.  Ans.  2x  —  3y  —  6  =  0. 

13.  What  is  the  locus  of  Ax  +  By  +  C  =  0  it  B  and  C  are  constant  and 
A  arbitrary  ?   if  -4  and  iJ  are  constant  and  C  arbitrary  ? 


92  ANALYTIC  GEOMETRY 

44.  Straight  lines  determined  by  two  conditions.  In  Ele- 
mentary Geometry  we  have  many  illustrations  of  the  determina- 
tion of  a  straight  line  by  two  conditions.  Thus  two  points 
determine  a  line,  and  through  a  given  point  one  line,  and  only 
one,  can  be  drawn  parallel  to  a  given  line.  Sometimes,  however, 
there  will  be  two  or  more  lines  satisfying  the  two  conditions ; 
thus  through  a  given  point  outside  of  a  circle  we  can  draw  two 
lines  tangent  to  the  circle,  and  four  lines  may  be  drawn  tangent 
to  two  circles  if  they  do  not  intersect. 

Analytically  such  facts  present  themselves  as  follows.  The 
equation  of  any  straight  line  is  of  the  form  (Theorem  II,  p.  86) 
(1)  Ax-\-mj-{-C  =  0, 

and  the  line  is  completely  determined  if  the  values  of  two  of  the 
coefficients  A,  B,  and  C  are  known  in  terms  of  the  third. 

For  example,  it  A  =  2B  and  C  —  —3B,  equation  (1)  becomes 
2Bz  +  Btj-3B==0, 
or  2x-\-y  —  3  =  0. 

Any  geometrical  condition  which  the  line  must  satisfy  gives 
rise  to  an  equation  between  one  or  more  of  the  coefficients 
A,  B,  and  C. 

Thus  if  the  line  is  to  pass  through  the  origin,  we  must  have  (7=0  (Theorem  VI, 

p.  73) ;  or  if  the  slope  is  to  he  3,  then =3  (Corollary  I,  p.  8G). 

B 

Two  conditions  which  the  line  must  satisfy  will  then  give  rise 
to  two  equations  in  A,  B,  and  C  from  which  the  values  of  two  of 
the  coefficients  may  be  determined  in  terms  of  the  third,  and  the 
line  is  then  determined. 

If  these  equations  are  of  the  first  degree,  there  will  be  only  one 
line  fulfilling  the  given  conditions,  for  two  equations  of  the  first 
degree  have,  in  general,  only  one  solution  (Theorem  IV,  p.  90). 
If  one  equation  is  a  quadratic  and  the  other  of  the  first  degree, 
then  there  will  be  two  lines  fulfilling  the  conditions,  provided 
that  the  solutions  of  the  equations  are  real.  And,  in  general, 
the  number  of  lines  fulfilling  the  two  given  conditions  will 
depend  on  the  degrees  of  the  equations  in  the  A,  B,  and  C  to 
which  they  give  rise.  ^ 


THE  STRAIGHT  LINE 


93 


I 


Rule  to  determine  the  equation  of  a  straight  line  which  satisfies 
two  conditions. 

First  step.    Assume  that  the  equation  of  the  line  is 
Ax -{- By  ^  C  =^  ^. 

Second  step.  Find  two  equations  between  A,  B,  and  C  each  of 
which  expresses  algebraically  the  fact  that  the  line  satisfies  one 
of  the  given  conditions. 

Third  step.  Solve  these  equations  for  two  of  the  coefficients  A, 
B,  and  C  in  tei^ms  of  the  third. 

Fourth  step.  Substitute  the  results  of  the  third  step  in  the  equct/- 
tion  in  the  first  step  and  divide  out  the  remaining  coefficient.  The 
result  is  the  required  equation. 

Ex.  1.  Find  the  equation  of  the  line  through  the  two  points  Pi  (5,  —  1) 
and  P2(2,  -2). 

Solution.    First  step.    Let  the  required  equation  be 

(1)  Ax  +  B2j-hC  =  0. 
Second  step.    Since  Pi  lies  on  the  locus 

of  (1)  (Corollary,  p.  53), 

(2)  6A-B+C  =  0', 
and  since  Pg  lies  on  the  line, 

(3)  2A-2B-i-C  =  0. 
Third  step.    Solving  (2)  and  (3)  for  A  and  B  in  terms  of  C,  we  obtain 

A=-IC,  B  =  IC. 
Fourth  step.    Substituting  in  (1), 

-lCx+lC2j  +  C  =  0. 
Dividing  by  C  and  simplifying,  the  required  equation  is 
x-3y-S  =  0. 

Ex.  2.  Find  the  equation  of  the  line,  passing  through  Pi  (3,  —  2)  whose 
slope  is  —  i. 

Solution.     First  step.    Let  the  re- 
quired equation  be 

(4)  Ax  +  By+C^O. 
Second  step.    Since  Pi  lies  on  (4), 

(5)  3A-2B+C  =  0; 
and  since  the  slope  is  —  i^ 

(6)  -^=-1. 


94  ANALYTIC  GEOMETRY 

Third  step.    Solving  (5)  and  (6)  for  A  and  C  in  terms  of  B,  we  obtain 

A  =  IB,  C  =  IB. 
Fourth  step.    Substituting  in  (4), 

lBx  +  By-^lB  =  0, 
or  a;  +  42/  +  5  =  0. 

PROBLEMS 

1.  Find  the  equation  of  the  line  satisfying  the  following  conditions  and 
plot  the  lines. 

(a)  Passing  through  (0,  0)  and  (8,  2). 

(b)  Passing  through  (-1,  1)  and  (-  3,  1). 

(c)  Passing  through  (—3,  1)  and  slope  =  2. 

(d)  Having  the  intercepts  a  =  o  and  &  =  —  2. 

(e)  Slope  =  —  3,  intercept  on  X-axis  =  4. 

(f )  Intercepts  a  =  —  3  and  6  =  —  4. 

(g)  Passing  through  (2,  3)  and  (-  2,  -  3). 
(h)  Passing  through  (3,  4)  and  (-  4,  -  3). 

(i)  Passing  through  (2,  3)  and  slope  =  —  2. 

(j)  Having  the  intercepts  2  and  —  5. 

2.  Find  the  equation  of  the  line  passing  through  the  origin  parallel  to  the 
line  2  cc  —  3  2/  =  4.  Ans.    2  x  —  3  ?/  =  0. 

3.  Find  the  equation  of  the  line  passing  through  the  origin  perpendicular 
to  the  line  5x4-2/  —  2  =  0.  Ans.   x  —  6y  =  0. 

4.  Find  the  equation  of  the  line  passing  through  the  point  (3,  2)  parallel 
to  the  line  4x  —  y  —  3  =  0.  Ans.    4x  —  y  —  10  =  0. 

5.  Find  the  equation  of  the  line  passing  through  the  point  (3,  0)  perpen- 
dicular to  the  line  2x  +  y  —  5  =  0.  Ans.   x  —  2y  —  3  =  0. 

6.  Find  the  equation  of  the  line  whose  intercept  on  the  Y-axis  is  5  which 
passes  through  the  point  (6,  3).  Ans.    x  -f  3y  —  15  =  0. 

7.  Find  the  equation  of  the  line  whose  intercept  on  the  JT-axis  is  3  which 
is  parallel  to  the  line  x  —  4?/-|-2  =  0.  Ans.    x  —  4?/  —  3  =  0. 

8.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  —  2y  -\-  S  =  0  and  x  +  2y  —  9  =  0. 

Ans.   X  —  y  =  0. 

9.  Find  the  equation  of  the  straight  line  whose  slope  is  m  which  passes 
through  the  point  Pi  (xi,  yi).  Ans.    y  —  yi  =  m{x  —  Xi). 


Ans. 

x-4y  =  0. 

Ans. 

y-i  =  o. 

An^. 

2  X  -  ?y  +  7  =  0. 

Ans. 

2x-3y  -6  =  0. 

Ans. 

3x-l-y-12  =0. 

Ans. 

4x  +  3?/-M2  =  0. 

Ans. 

Sx-2y  =  0. 

Ans. 

x-y  +  1  =  0. 

Ans. 

2x  +  y-l  =  0. 

Ans. 

^--y-  =  i. 

2      5 

THE  STRAIGHT  LINE  95 

10.  Find  the  equation  of  the  straight  line  whose  intercepts  are  a  and  b. 

Ans.    -  +  l  =  l. 
a      0 

11.  Find  the  equation  of  the  straight  line  passing  through  the  points 

Pi{xu  y\)  and  Pa  (0^2,  2/2). 

,  Ans.   (2/2  -Vi)^-  {X2  -Xi)y  +  x^yi  -  xiy^  =  0. 

12.  Show  that  the  result  of  the  last  problem  may  be  put  in  the  form 

x-xi  ^y-yi 
X2-X1     y^-yi 

Hint.    Add  and  subtract  x^y^,  factor,  transpose,  and  express  as  a  proportion. 

45.  The  equation  of  the  straight  line  in  terms  of  its  slope 
and  the  coordinates  of  any  point  on  the  line.  In  this  section 
and  in  those  immediately  following,  the  Rule  in  the  preceding 
section  is  applied  to  the  determination  of  general  forms  of  the 
equations  of  straight  lines  satisfying  pairs  of  conditions  which 
occur  frequently.  These  general  forms  will  then  enable  us  to 
write  the  equations  of  certain  straight  lines  with  the  same  ease 
that  the  equation  y  —  mx  +  h  enables  us  to  write  the  equation 
of  the  straight  line  whose  slope  and  intercept  on  the  F-axis  are 
given. 

Theorem  V.  Point-slope  form.  The  equation  of  the  straight  line 
which  passes  through  the  point  Pi  (xi,  y^  and  has  the  slope  m  is 

(V)      -  y  -.y^z=m{pc  —  cCi). 

Proof.    First  step.    Let  the  equation  of  the  given  line  be 

(1)  Ax-{-By  -\-C  =^0. 

Second  step.    Then,  by  hypothesis, 

(2)  Axi  +  5?/i  +  C  =  0 
and 

(3)  -f  =  ^- 

Third  step.  Solving  (2)  and  (3)  for  A  and  C  in  terms  of  By 
we  obtain 

A  =—  mB  and  C  =  B (inxi  —  y^). 


96  ANALYTIC  GEOMETRY 

Fourth  step.    Substituting  in  (1),  we  have 

—  mBx  -\-  By  -\-  B  (^mx-i  —  y^)  =  0. 
Dividing  by  B  and  transposing, 

y  —  yi  =  m{x  —  x^).  ^      q.e.b. 

If  Pi  lies  on  the  T-axis,  cci  =  0  and  y^  =  b,  so  that  this  equa- 
tion becomes  y  =  mx  +  b. 

46.  The  equation  of  the  straight  line  in  terms  of  its  intercepts. 

We  pass  now  to  the  consideration  of  a  line  determined  by  two 
points,  and  we  consider  first  the  case  in  which  the  two  points  lie 
on  the  axes.  This  section  does  not,  therefore,  apply  to  lines  par- 
allel to  one  of  the  axes  or  to  lines  passing  through  the  origin,  as  in 
the  latter  case  the  two  points  coincide  and  hence  do  not  deter- 
mine a  line. 

Theorem  VI.    Intercept  form.    If  a  and  b  are  the  intercepts  of  a  line 
on  the  X-  and  Y-axes  7'espectively,  then  the  equation  of  the  line  is 

(VI)  -  +  1^  =  1. 

^      ^  ah 

Proof     First  step.    Let  the  equation  of  the  given  line  be 

(1)  Ax  +  By  +  C  =  0. 

Second  step.    By  definition  of  the  intercepts  (p.  73),  the  points 
(a,  0)  and  (0,  b)  lie  on  the  line;  hence 

(2)  ^a  +  <^  =  0, 

(3)  Bb-\-C  =  0. 

Third  step.    Solving  (2)  and  (3)  for  A  ar«d  B  in  terms  of  C. 

we  obtain 

A  = C  and  B  =  --C. 

a  0 

Fourth  step.    Substituting  in  (1),  we  have 

-  -  Cx  -  -  Cy  +  C  =  0. 
a  b 

Dividing  by  C  and  transposing, 

^   ,   ?/  _  1 

r  T  —  -L-  Q.E.D. 


THE  STRAIGHT  LINE 


97 


Ex.  1.    Write  the  equation  of  the  locus  of  2x  —  62/  +  3  =  0in  terms  of 
its  intercepts  and  plot  the  line. 

Solution.     Transposing  the  constant  term,  we  have 

Dividing  by  —  3, 
2x 
-3 

X 


+  2y  =  l, 


_.  +  ?  =  >■ 

2  2 

This  equation  is  of  the  form  (VI).     Hence 
a  =  —  I  and  b  =  |. 
Plotting  the  points  (—  |,  0)  and  (0,  I)  and  joining  them  by  a  straight  line, 
we  have  the  required  line. 

47.  The  equation  of  the  straight  line  passing  through  two 
given  points. 

Theorem  VII.  Two-point  form.  The  equation  of  the  straight  line 
passing  through  Pi  (xi,  y{)  and  P^,  (x^,  ])%)  is 

/yil)  ^  -  ^1  ^  y  -  Z/1 

^       ^  a?2  —  a?i       2/2  —  2/i 

Proof.    Let  the  equation  of  the  line  be 

(1)  Ax-{-By^-C  =  0. 

Then,  by  hypothesis, 

(2)  Axi  +  By,  +  C  =  0 
and 

(3)  Ax^  +  %2  +  C  =  0. 

To  follow  the  Rule,  p.  93,  we  must  solve  (2)  and  (3)  for  A 
and  B  in  terms  of  C,  substitute  in  (1),  and  divide  by  C ;  that  pro- 
cedure amounts  to  eliminating  A,  B,  and  C  from  (1),  (2),  and  (3), 
and  that  elimination  may  be  more  conveniently  performed  as 
follows : 

Subtract  (2)  from  (1)  ;  this  gives 

A{x-x,)+B{y-y,)=0, 


or 
(4) 


A{x-a:,)  =  -B{y-yi). 


98  ANALYTIC  GEOMETKY 

Similarly,  subtracting  (2)  from  (3),  we  obtain 
(5)  A  (x,  -x,)  =  -B  (2/2  -  2/i). 

Dividing  (4)  by  (5),  we  find 

= Q.E.D. 

*^2     "^i     2^2     yi 

Corollary.  The  condition  that  three  points,  Pi(xi,  i/i),  ^2(^2?  2/2)? 
and  Pg  (ccg,  2/3)  should  lie  on  a  line  is  that 

^z  —  ^i^yz  —  yi 
^2  —  ^1     2/2     yi 

For  this  is  the  condition  that  P3  should  lie  on  the  line  (VII)  passing  through 
i^jj^ndL>P2  (Corollaiy,  p.  53). 

The  TYiethod  of  proving  the  corollary  should  be  remembered 
rather  than  the  corollary  itself,  as  then  the  condition  may  be 
immediately  written  down 'from  (VII). 

PROBLEMS 

1 .  Find,  by  substitution  in  the  proper  formulas,  the  equations  of  the  lines 
satisfying  the  conditions  in  problem  1,  p.  94. 

2.  Find  the  equations  of  the  lines  fulfilling  the  following  conditions  and 
plot  the  lines. 

(a)  Passing  through  the  origin,  slope  =  3.  Ans.  Sx  —  y  =  0. 

(b)  Passing  through  (3,  -  2)  and  (0,  -  1).  Ans.  x  +  3?/  +  3  =  0. 

(c)  Having  the  intercepts  4  and  —  3.          '  Ans.  Sx  —  4:y  —  12  =  0. 

(d)  F-intercept  =  5  and  slope  =  3.  Ans.  3x  —  y  +  5  =  0. 

(e)  Passing  tiirough  (1,  —  2)  and  (3,  —  4).  Ans.  x  +  y  +  1  =  0. 

(f )  Having  the  intercepts  —  1  and  —  3.  Ans.  Sx-\-y  +  3  =  0. 

(g)  Passing  through  (-  |,  |)  and  slope  =  -  f .       Ans.  ix  +  6y—7  =  0. 
(h)  Passing  through  (0,  0)  and  slope  =  m.  Ans.  y  =  mx. 

/T'        3.  Find  the  equations  of  the  sides  of  the  triangle  whose  vertices  are 
(-3,2),  (3,  -  2),  and  (0,  -  1). 

Ans.    2x  +  32/  =  0,  a;  +  3?/  +  3  =  0,  and  x  +  ?/  +  1  =  0. 

4.  Find  the  equations  of  the  medians  of  the  triangle  in  problem  3  and 
show  that  they  meet  in  a  point. 

Ans.   x  =  0,7x-\-9y  -\-  S  =  0,  and  5x  +  9y  -{■  S  =  0. 

Hint.  To  show  that  three  lines  meet  in  a  point,  find  the  point  of  intersection  of  two 
of  them  and  prove  that  it  lies  on  the  third. 


THE  STRAIGHT  LINE  99 

5.  Show  that  the  medians  of  any  triangle  meet  in  a  point. 

Hint.    Taking  one  vertex  for  origin  and  one  side  for  the  X-axis,  the  vertices  may  then 
be  called  (0,  0),  (a,  0),  and  (6,  c). 

6.  Determine  whether  or  not  the  following  sets  of  points  lie  on  a  straight 
line. 

(a)  (0,  0),  (1,  1),  (7,  7).  Ans.  Yes. 

(b)  (2,  3,),  (-  4,  -  6),  (8,  12).  Ans.  Yes. 

(c)  (3,  4),  (1,  2),  (5,  1).  Ans.  No. 

(d)  (3,  -  1),  (-  6,  2),  (-  I,  1).  Ans.  No. 

(e)  (5,  6),  (1,1),  (-1,-1).  ^ns.  Yes. 

(f)  (7,  6),  (2,  1),  (6,  -  2).  Ans.  No. 

7.  Reduce  the  following  equations  to  the  form  (VI)  and  plot  their  loci. 

(a)  2x-f  32/-6  =  0.  (d)  3x  +  4y  +  1  =  0. 

(b)  x-32/  +  6  =  0.  (e)  2x-4y-7=0. 

(c)  3x-42/  +  9  =  0.  (f)  7x-6y-Z  =  0. 

8.  Find  the  equations  of  the  lines  joining  the  middle  points  of  the  sides 
of  the  triangle  in  problem  3  and  show  that  they  are  parallel  to  the  sides. 

•   Ans.    4x  +  6?/  +  3  =  0,  x  +  3?/=:0,  and  x-\-y  =  0. 

9.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  +  2  y  =  1  and  2x-4?/-3  =  0. 

Ans.   X  +  10  y  =  0. 

10.  Show  that  the  diagonals  of  a  square  are  perpendicular. 
Hint.    Take  two  sides  for  the  axes  and  let  the  length  of  a  side  be  a. 

11.  Show  that  the  line  joining  the  middle  points  of  two  sides  of  a  triangle 
is  parallel  to  the  third. 

Hint.    Choose  the  axes  so  that  the  vertices  are  (0,  0),  (a,  0),  and  (6,  c). 

12.  Find  the  equation  of  the  line  passing  through  the  point  (3,  -  4)  which 
has  the  same  slope  as  the  line  2 x  -  y  =  3.  Ans.   2x-y  -10  =  0. 

13.  Find  the  equation  of  the  line  passing  through  the  point  (-1,  4)  which 
is  parallel  to  the  line  3 x  +  y  +  1  =  0.  Ans.   3x  +  y  -1  =  0. 

14.  Two  sides  of  a  parallelogram  are2x  +  3y-7  =  0  and  x-3y  +  4  =  0. 
Find  the  other  two  sides  if  one  vertex  is  the  point  (3,  2). 

Ans.   2x-f3y- 12  =  0  and  x-3y  +  3  =  0. 

15.  Find  the  equation  of  the  line  passing  through  the  point  (- 2,  3) 
which  is  perpendicular  to  the  line  x-{-2y  =  1.       Ans.    2x  -y  +  1  =  0. 


100  ANALYTIC  GEOMETRY 

16.  Show  that  the  three  lines  x  —  2y  =  0,  x-{-2y  —  8  =  0,  and  x  +  2y 
—  S  +  k{x  —  2  y)  =  0  meet  in  a  pomt  no  matter  what  value  k  has,  ' 

17.  Derive  (V)  and  (VII)  by  the  Rule  on  p.  53,  using  Theorem  V,' p.  35. 

18.  Derive  (VI)  and  (VII)  by  the  Rule  on  p.  53,  using  the  theorem  that 
the  corresponding  sides  of  similar  triangles  are  proportional. 

19.  Derive  y  =  mx  +  h  and  (V)  by  the  Rule  on  p.  53,  using  the  definition 
of  the  tangent  of  an  acute  angle  in  a  right  triangle.  ' 

20.  Derive  the  equation  of  the  straight  line  in  terms  of  the  perpendicular 
distance  p  from  the  origin  to  the  line  and  the  angle  w  which 
that  perpendicular  makes  with  the  positive  direction  of 
the  X-axis. 

Hint.    Find  the  intercepts  in  terms  of  p  and  w  by  solving  the 
right  triangles  in  the  figure  and  substitute  in  (VI). 

Ans.   X  cos  w  +  2/  sin  w  —  p  =  0. 


V^ 


21.  What  is  the  locus  of  (V)  if  Xi  and  2/1  are  constant  and  m  arbitrary  ? 

22.  What  is  the  locus  of  (VI)  if  a  is  constant  and  6  arbitrary  ?  if  6  is  con- 
stant and  a  arbitrary  ?  ,.,j^- ' 

23.  Write  an  equation  which  represents  all  lines  passing  through  (2,  —  1). 

24.  Write  an  equation  representing  all  lines  whose  intercept  on  the  X-axis 
is  3. 

25.  Write  in  two  different  forms  the  equation  of  all  lines  whose  intercept 
on  the  Y-axis  is  —  2. 

26.  Write  an  equation  representing  all  lines  whose  slope  is  —  \. 

27.  If  the  axes  are  oblique  and  make  an  angle  of  w,  then  the  equation  of  a 
straight  line  in  terms  of  its  inclination  a  and  intercept  on  the  Y-axis  h  is 

sin  a 


sin  (w  —  a) 


«  +  &. 


28.  If  the  angle  between  the  axes  is  w,  the  equation  of  the  line  passing 
through  Pi(iCi,  2/1)  whose  inclination  is  a  is 

sin  a       ,  , 

y  —  y^  —  - — ^^  {x  —  Xx). 

-^      ^^      sm  (w  -  a)  ^  '^ 

29.  Show  that  equations  (VI)  and  (VII)  hold  for  obhque  coordina-tes. 


ril 


THE  STRAIGHT  LINE 


101 


48.  The  normal  form  of  the  equation  of  the  straight  line. 

In  the  preceding  sections  the  lines  considered  were  determined 
by  two  points  or  by  a  point  and  a  direction.  Both  of  these 
methods  of  determining  a  line  are  frequently  used  in  Elementary 
Geometry,  but  we  have  now  to  consider  a  line  as  determined  by 
two  conditions  which  belong  essentially  to  Analytic  Geometry. 


Y 

AV 

A 

r\. 

0 

> 

'  X 

<^ 

7- 

B 

•^ 

Ox 

^^ 

^A 

A 

Vx 

I't           ,A 

& 

^-v 

/  > 

M 

Let  ABhe  any  line,  and  let  ON  be  drawn  from  the  origin  perpen- 
dicular to  J. ^  at  C.  Let  the  positive  direction  on  ON  he  from  0 
toward  N,  —  that  is,  from  the  origin  toward  the  line,  —  and  denote 
the  positive  directed  length  OC  by  ^  and  the  positive  angle 
XON,  measured,  as  in  Trigonometry  (p.  18),  from  OX  as  initial 
line  to  ON  as  terminal  line,  by  (o*  Then  it  is  evident  from  the 
figures  that  the  position  of  any  line  is  determined  hy  a  pair  of 
values  of  p  and  oo,  both  p  and  w  being  positive  and  w  <  2  tt. 

On  the,  other  hand,  every  line  determines   a  single  positive 
value  of  p  and  a  single  positive  value  of  a>  which  is  less  than 


i\^\ 

Y> 

' 

\ 

^ 

^A 

^ 

A 

\ 

X 

n 

\ 

-K-rf 

2  7r,  unless  ^  =  0.  When  p  =  0,  however,  AB  passes  through 
the  origin,  and  the  rule  given  above  for  the  positive  direction 
on  ON  becomes  meaningless.  From  the  figures  we  see  that  we 
can  choose  for  <u  either  of  the  angles  XON  or  XON'.  When 
p  —  O  we  shall  always  suppose  that  oiKir  and  that  the  positive 
direction  on  ON  is  the  upward  direction. 

*  ui  is  not  the  angle  between  the  directe^J  lines  OX  and  OX,  as  defined  on  p.  28. 


102  ANALYTIC  GEOMETRY 

Theorem  VIII.    The  normal  form*  of  the  equation  of  the  straight 
line  is 
(YIII)        •  oc  cos  (H  -\-  y  siiKH  —  p  =  O, 

where  p  is  the  perpendicular  distance  or  normal  from  the  origin  to 
the  line  and  w  is  the  positive  angle  which  that  perpendicular  makes 
with  the  positive  direction  OX  of  the  X-axis  regarded  as  initial 
line. 

Proof    Let  P{x,  y)  be  any  point  on  the  given  line  AB. 

Then  since  AB  is  perpendicular  to 
ON,  the  projection  of  OP  on  ON  is 
equal  to  p  (definition,  p.  29).  By 
the  second  theorem  of  projection 
(p.  48),  the  projection  of  OP  on  ON 
is  equal  to  the  sum  of  the  projections 
^  ^  of  OD  and  DP  on  ON.  Then  the  con- 
dition that  P  lies  on  ^jB  is 

(1)  pfoj.  of  OD  on  ON  -f  proj.  of  DP  on  ON  =  p. 
By  the  first  theorem  of  projection  (p.  30)  we  have 

(2)  pi'oj-  of  OD  on  ON  =  OD  cos  w  —  x  cos  o>, 

(3)  proj.  of  DP  on  ON  =  DP  cos  (  —  —  w  J  =  ?/  sin  w. 

For  the  angle  between  the  directed  lines  DP  and  ON  equals  that  between 
OrandOiV=|-a;. 

Substituting  from  (2)  and  (3)  in  (1),  we  obtain 

£c  cos  0)  +  y  sin  o)  — ^  =  0.  q.e.d. 

To  reduce  a  given  equation 

(4)  Ax+By  +  C^O 

to  the  normal  form,  we  must  determine  w  and  p  so  that  the  locus 
of  (4)  is  identical  with  the  locus  of 

(5)  X  cos  0)  +  2/  sin  CO  —  ^  =  0. 

*  The  designation  of  this  equation  is  made  clear  by  the  definition  of  the  normal  in 
Chapter  IX. 


;3 


THE  STRAIGHT  LINE  103 

Then  we  must  have  corresponding  coefficients  proportional 
(Theorem  III,  p.  88). 

cos  0)  _  sin  (n  _—  p 
•'•     A     ~     B     ~  ~C" 

Denote  the  common  value  of  these  ratios  by  r ;  then 

(6)  cos  o>  =  rA, 

(7)  sin  CD  =  rB,  and 

(8)  -p  =  rC. 

To  find  r,  square  (6)  and  (7)  and  add ;  this  gives 

sin2  io  -f  cos^  to  =  r2(^2  _^  B""). 
But  sin^  w  4-  cos^  w  =  1 ; 

and  hence  r^(^^  +  B"^)  =  1,  or 

(9)  r  = ■  ^ 

Equation  (8)  shows  which  sign  of  the  radical  to  use ;  for  since 
p  is  positive,  r  and  C  must  have  opposite  signs,  unless  C  =  0.  If 
C  =  0,  then,  from  (8),  ^  =  0,  and  hence  w  <  tt  (p.  101) ;  then  sin  a> 
is  positive,  and  from  (7)  r  and  B  must  have  the  same  signs. 

Substituting  the  value  of  r  from  (9)  in  (6),  (7),  and  (8)  gives 

A                 .                      B  C 

cos  (0  = ■  ?    sm  0)  = ,  ?   p  = ■ 

Hence  (5)  becomes 

(10)     /        •  x  + y  + ,         .      =  0, 

^   ^    ±VlMr^      ±-y/A^-\-B^      iVI^:^ 

which  is  the  normal  form  of  (4).     The  result  of  the  discussion 
may  be  stated  in  the  following 

Rule  to  reduce  Ax  -\-  By  -\-  C  =  0  to  the  normal  form. 

First  step.    Find  the  numerical  value  of  V^  ^  -f  B"^. 

Second  step.  Give  the  result  of  the  first  step  the  sign  opposite  to 
that  of  C,  or,  if  C  =  0,  the  same  sign  as  that  of  B. 

Third  step.  Divide  the  given  equationlby  the  result  of  the  second 
step.     The  result  is  the  required  equation. 

I 


104  ANALYTIC  GEOMETRY 

The  advantages  of  the  normal  form  of  the  equation  of  the 
straight  line  over  the  other  forms  are  twofold.  In  the  first 
place,  every  line  may  have  its  equation  in  the  normal  form; 
whether  it  is  parallel  to  one  of  the  axes  or  passes  through  the 
origin  is  immaterial.  In  the  second  place,  as  will  be  seen  in  the 
following  section,  it  enables  us  to  find  immediately  the  distance 
from  a  line  to  a  point. 


PROBLEMS 

\ 

1.  In  what  quadrant  will  ON  (Fig.,  p.  101)  lie  if  sin  w  and  cos  w  are  both 
positive?  both  negative?  if  sinw  is  positive  and  cosw  negative?  if  sinw 
is  negative  and  cos  w  positive  ?  ^ 

2.  Find  the  equations  and  plot  the  lines  for  which 
(a)  w  =  0,  p  =  5.  Arts,   x  =±  ^. 

Atis.   ?/  +  3  =  0. 

Ans.    V2x+ V2?/- 6  =  0. 

Ans.   ic— V32/  +  4  =  0. 

7  Tf 

i\^      (e)  w  =  -— ,  j9  =  4.  Ans.    V2 x  -  V2y  -8  =  0. 

3.  Reduce  the  following  equations  to  the  normal  form  and  find  p  and  w. 

(a)  3a;  +  42/-2  =  0.  Ans.  p  =  |,  w  =  cos-i  f  =  sin- 1  f . 

(b)  3x  -  4y  -^  2  =  0.  Ans.  p  =  f,  a>  =  cos-i  |  =  sin-i (-  f). 

(c)  12  a;  -  5  ?/  =  0,  Ans.  p  =  0,  w  =  cos-i  (-  ff)  =  sin-^i  j-^. 

(d)  2x  +  5?/  +  7  =  0.  ^ 

Ans.   p  = — ,  a;  =  cos-i(' — ^"j  ^z  sin- Y ~ Y 

+  V2y  \_V29/  V-V29^ 

(e)  4aj  -  3?/  +  1  =  0.  Ans.   p  =  1,  w  =  cos-i(-  ij  =  sin-if. 

(f)  4x-5  2/  +  6  =  0. 


D(b)  c-. 

2 

,  p 

=  3. 

0    .(c)  0,. 

7t 

p  = 

3. 

;..,(d)  0,. 

_27t 

'  p 

=  2. 

Ans.   p  = 


\  _  V41  /  \  a- VIT/ 


Vii  V_V4i/  WV41 


4.  Find  the  perpendicular  distance  from  the  origin  to  each  of  the  follow- 
ing lines. 

(a)  12x+5?/-26  =  0.  Ans.    2. 

(b)  x-\-y  +  l=0.  Ans.    1V2. 

(c)  3x-2?/-l  =  0.  Ans.    j\^Ts. 


ri' 


THE  STRAIGHT  LINE 


105 


5.  Derive  (VIII)  when  (a)  -<w<7r;   (b)  n<(a< — ;  (c)  ^ — <w<2^; 
(d)  p  =  OandO<w<-. 

6.  For  what  values  of  p  and  w  will  the  locus  of  (VIII)  be  parallel  to  the 
X-axis  ?   the  F-axis  ?   pass  through  the  origin  ? 

7.  Find  the  equations  of  the  lines  whose  slopes  equal  —  2,  which  are  at  a 
distance  of  5  from  the  origin, 

Ans.    2V5X+ VSy- 25  =  0  and  2V5x+V5?/  + 25  =  0. 

8.  Find  the  lines  whose  distance  from  the  origin  is  10,  which  pass  through 
the  point  (5,  10).  Ans.   y  =  10  and  4  x  +  3  y  =  50.     J 

9.  "What  is  the  locus  of  (VIII)  if  p  is  constant  and  w  arbitrary  ?   if  w  is 
constant  and  p  arbitrary  ? 

10.  Write  an  equation  representing  all  lines  whose  distance  from  the 
origin  is  5. 

49.  The  distance  from  a  line  to  a  point.  The  positive  direction 
on  the  normal  ON  drawn  through  the  origin  perpendicular  to  AB 
(Fig.  1)  is  from  0  to  AB  (p.  101) ;  and  when  AB  passes  through  0 
(Fig.  2)  the  positive  direction  on  ON  is  the  upward  direction. 


A 

yN 

i\, 

/ 

i . 

JC' 

0/ 

^ 

(2) 

The  positive  direction  on  ON  is  taken  to  be  the  positive  direction 
on  all  lines  perpendicular  to  AB.  Hence  the  distance  from  the 
line  AB  to  the  point  Pj  is  positive  if  P^  (ind  the  origin  are  on 
op2Josite  sides  of  AB,  and  negative  if  Pi  and  the  origin  are  on  the 
same  side  of  AB.  When  A B  passes  through  the  origin  the  distance 
from  A  B  to  Pi  is  positive  if  that  distance  is  in  the  upward,  direc- 
tion, and  negative  if  it  is  in  the  doivnumrd  direction.  Thus  in  the 
figures  the  distance  from  AB  to  Pi  is  positive  and  from  AB  to  P^ 
is  negative. 


106 


ANALYTIC  GEOMETRY 


Theorem  EX.    The  distance  d  from  the  line 
X  cos  (0  +  2/  sin  to  —  ^  =  0 
to  the  point  P^  (iCj,  ?/i)  is 
(IX)  •  e^  =  a?i  cos  (0  -f  ^1  sin  (H  —  p. 

Proof.    Let  AB  he  the  given  line  and  let  ON  be  perpendicular 
to  AB.     By  the  second  theorem  of  projection  (p.  48)  we  have 

proj.  of  OPx  on  ON  =  proj.  of  OD  on  ON  +  proj.  of  Dl^^  on  ON. 

From  the  figure, 
proj.  of  OPx  on  ON 

=  OE=p  +  d. 
By   the    first   theorem   of 
projection  (p.  30), 
proj.  of  OD  on  ON 

=  OD  cos  (0  =  iCi  cos  ft), 
proj.  of  DPi  on  ON 

TT 


=  i)Pi  COS 

=  2/i  sin  (o. 

Hence  p  ■\-  d  =  x^  cos  w  +  ?/i  siri  co, 

and  therefore  d  =  x^  cos  w  +  ?/i  sin  la  —  p. 

Erom  this  theorem  we  have  at  once  the 


-) 


Q.E.D. 


Rule  to  find  the  perpendicular  distance  from  a  given  line  to  a 
given  point. 

First  step.  Beduce  the  equation  of  the  given  line  to  the  normal 
form  {Rule,  p.  103). 

Second  step.  Substitute  the  coordinates  of  the  given  point  for 
X  and  y  in  the  left-hand  side  of  the  equation.  The  result  is  the 
required  distance. 

The  sign  of  the  result  will  show  on  which  side  of  the  line  the 
point  lies. 


THE  STRAIGHT  LINE 


107 


Ex.  1.    Find  the  distance  from  the  line  4ic  —  3y  +  15  =  0  to  the  point 
(2,  1). 

Fa 

Solution.    First  step.   Reducing  the  given  equation 

to  normal  form,  we  have 

-|x  +  |y-3  =  0. 

Second  step.     Substituting  2  for  x  and  1  for  y, 
we  have 

^=-f-2  +  f(l)-3=-4. 

What  does  the  negative  sign  mean  ? 


— 

— ' 

^y 

— 

— 1 

— 

/ 

y 

f 

/ 

s 

< 

/ 

V 

s 

/ 

3' 

x^ 

(\ 

) 

y 

0 

i 

_ 

Ex.  2.  Prove  that  the  sum  of  the  distances  from  the  legs  of  an  isosceles 
triangle  to  any  point  in  the  base  is  constant. 

Solution.  Take  the  middle  point  of  the  base  for  origin  and  the  base  itself 
for  the  X-axis.  Then  the  values  of  p  for  the  two  legs  are  equal  and  the  values 
of  (1)  are  supplementary.     Hence,  if  the  equation  y 

of  one  leg  in  normal  form  is 

X  cos  a>  +  2/  sin  w  —  p  =  0, 
then  the  equation  of  the  other  leg  is 

X  cos  (tt  —  co)  +  y  sin  (tt  —  w)  —  _p  =  0, 
or  —  X  cos  (o  +  y  sin  (a  —  p  =  0. 

Let  (a,  0)  be  any  point  in  the  base.  Then  the  distances  from  the  legs  to 
(a,  0)  are  respectively  a  cos  w  —  _p  and  —  a  cos  w  —  p,  so  that  the  sum  of  these 
distances  is  —  2  _p,  that  is,  a  constant. 


PROBLEMS 

1 .  Find  the  distance  from  the  line 

(a)  xcos45°  + 2/sin45°- V2  =  0  to  (5,  - 

(b)  |x-fy-l  =  0  to  (2,1). 

(c)  3x  +  4y  +  15  =  0  to  (-2,  3). 


7). 


(d)  2x-7y  +  8  =  0  to  (3, 

(e)  x-Sy  =  0  to  (0,4). 


5). 


Ans. 

-2V2. 

Ans. 
Ans. 

Ans. 

49 

Ans. 


+  V53 
12 


+  V10 


2.  Do  the  origin  and  the  point  (3,  -  2)  lie  on  the  same  side  of  the  line 
x-2/  +  l  =  0?  Ans.    Yes.    ' 


108  ANALYTIC   GEOMETKY 

3.  Does  the  line  2x  +  3?/  +  2  =  0  pass  between  the  origin  and  the  point 
(-2,  3)?  Ans.    No. 

4.  Pind  the  lengtlis  of  the  altitudes  of  the  triangle  formed  by  the  lines 
2x  +  3y  =  0,  x  +  3^  +  3  =  0,  and  x  +  y  +  1  -  Q. 

Ans.    — ^,  — =:,  and  V2. 
V13    Vio 

5.  Find   the   distance  from    the   line  Ax  -{-  By  -{-  G  =  0   to   the  point 
Pi(xi,  yi).  ■  ^^    Axi  +  Byi+C 

6.  Prove  Theorem  IX  when 

/  ^  n  tt,,  ,7r^  .  ^  S  7t     , ,.  S  7t  - 

(a)p  =  0,  a;<--;   (b)-<a;<7r;   (c)  tT  <  a,  < -- ;   (d) -— <  w  <  2  tT. 

^_       7.  Eind  the  locus  of  all  points  which  are  equally  distant  from 
3x-4y  +  l=:0  and  4x  +  3?/-l=0.    . 

Auii.    7x  —  y  =  0  and  x  +  7y  —  2  =  0. 

8.  Find  the  locus  of  all  points  which  are  twice  as  far  from  the  line 
12 X  +  6  y  —  1  =  0  as  from  the  F-axis.  Ans.    14  x  —  6  ?/  +  1  =  0. 

9.  Find  the  locus  of  points  which  are  k  times  as  far  from  Ax  —  Sy  +1=0 
as  from  5x  -  12?/  =  0.  Ans.   (52  -  25fc)x  -  (39  -  60k)y  +  13  =  0. 

10.  Find  the  bisectors  of  the  angles  formed  by  the  lines  in  problem  9. 

Ans.    77x -99?/  + 13  =  0  and  27x  +  21?/  + 13  =  0. 

11.  Find  the  distance  between  the  parallel  lines, 
3x  +  l,       ,  3  ,^,    r?/  =  mx  +  3,      ,  6 


<^){:::3::i:  --^  <^>{. 


+  VlO  Ly  =  mx-3.  +VTT^ 

12.  Derive  the  normal  equation  of  the  line  by  means  of  Theorem  IX. 

13.  Prove  that  the  altitudes  on  the  legs  of  an  isosceles  triangle  are  equal. 

14.  Prove  that  the  three  altitudes  of  an  equilateral  triangle  are  equal. 

15.  Prove  that  the  sum  of  the  distances  from  the  sides  of  an  equilateral 
triangle  to  any  point  is  constant. 

Hint.    Take  the  center  of  the  triangle  for  origin,  with  the  JC-axis  parallel  to  one  side. 


THE  STRAIGHT  LINE 


109 


fO 


16.  Find  the  areas  of  the  triangles  formed  by  the  following  lines. 

(a)  2  X  -  3  ?/  +  30  =  0,  X  =  0,  X  +  2/  =  0.  Ans.  30. 

(b)  x  +  2/  =  2,  3x  +  4y- 12  =  0,  x-?/+6  =  0.  Ans.  f 

(c)  3x  -  4y  +  12  =  0,  X  -  3?/  +  6  =  0,  2x  -  ?/  =  0.  Am.  33. 

(d)  x  +  3y-3  =  0,  5x-2/-15  =  0,  x-^+l  =  0.  Ans.  8.' 

17.  Plot  the  following  lines  and  find  the  area  of  the  quadrilaterals  of 
which  they  are  the  sides. 

(a)  X  =  ?/,  y  ==  6,  X  +  y  =  0,  3x  +  2 ?/  -  6  =  0.  Ans.  16i. 

(b)x  +  22/-5  =  0,  2/  =  0,  x  +  42/  +  5  =  0,  2x  +  ?/-4  =  0.    Ans.  18. 
(c)2x-42/-f8  =  0,  x  +  y  =  0,  2x- 2/^4  =  0,  2x4-y-3  =  0. 

Ans.  4tVo. 

50.  The  angle  which  a  line  makes  with  a  second  line.     The 

angle  between  two  directed  lines  has  been  defined  (p.  28)  as  the 
angle  between  their  positive  directions.  When  a  line  is  given 
by  means  of  its  equation,  no  positive  direction  along  the  line  is 
fixed.  In  order  to  distinguish  between  the  two  pairs  of  equal 
angles  which  two  intersecting  lines  make  with  each  other  we 
define  the  angle  which  a  line  makes  with  a 
second  line  to  be  the  positive  angle  (p.  18) 
from  the  second  line  to  the  Jii^st  line. 

Thus  the  angle  which  Li  makes  with  L^ 
is  the  angle  0.  We  speak  always  of  the 
'■'  angle  which  one  line  makes  with  a  second 
line,"  and  the  use  of  the  phrase  "  the  angle 
between  two  lines  "  should  be  avoided  if  those 
lines  are  not  directed  lines.  We  have  thus  added  a  third  method 
of  designating  angles  to  those  given  on  p.  18  and  p.  28. 

;  Theorem  X.    The  angle  6  which  the  line 

L,  :  A,x  ^B,7j  +  C,  =  0 
makes  with  the  line 


is  given  hy 
(X) 


tan^  = 


AxA2  +  B1JB2 


110 


ANALYTIC  GEOMt:TRY 


Proof.  Let  a-^  and  a^  be  the  inclinations  of  L^  and  L^  respec- 
tively. Then,  since  the  exterior  angle  of  a  triangle  equals  the 
sum  of  the  two  opposite  interior  angles,  we  have 

In  Fig.  1,  ai  =  $  -\-  a^,  or  6  =  a^  —  a^, 

In  Fig.  2,  a2  =  'Tr  —  6  +  ai,  or  0  =  tt  +  (ai  —  a^). 


And  since  (5,  p.  20) 

tan  (tt  -\-  <f>)  =  tan  <^, 
we  have,  in  either  case, 

tan  0  =  tan  (a^  —  a^) 

tan  ai  —  tan  org 


1  +  tan  ai  tan  a^ 


(by  13,  p.  20) 


But  tan  ai  is  the  slope  of  Zj  and  tan  ^2  is  the  slope  of  La ;  hence 
(Corollary  I,  p.  86) 


tan  0 


Reducing,  we  get   tan  0 


B,       B, 


-( 

:-^)( 

-I-) 

A,B, 

-  A,B, 

A^A^  +  B^B^ 


Q.E.D. 


Corollary.     If  mi  and  m^  are  the  slopes  of  two  lines,  then  the 
angle  $  which  the  first  line  makes  with  the  second  is  given  by 


tan^  = 


1  +  minii 


THE  STRAIGHT  LINE 


111 


Ex.  1.    Find  the  angles  of  the  triangle  formed  by  the  lines  whose  equations 
are 

L:2x-Sy-6  =  0, 

M:6x-y-6  =  0, 

N:6x  +  iy  -25  =  0. 

Solution.  To  see  which  angles  formed  by  the  given 
lines  are  the  angles  of  the  triangle,  we  plot  the  lines, 
obtaining  the  triangle  ABC.  A  is  the  angle  which 
M  makes  with  X,  so  that  M  takes  the  place  of  Li  in 
Theorem  X  and  L  of  Lz- 
Hence 

Ai  =  6,  Bi=-1; 
^2  =  2,  B2=-3. 


Then 

tan  1  -  ^'^^  ~  ^^^'  - 
ArA2  +  B1B2 

-2  +  18 
12  +  3 

16 
"15 

and  hence 

^  =  tan-i(H). 

B  is  the  angle  which  L  makes  with  N,  and  by  Corollary  III,  p.  87,  5  =  —  • 
C  is  the  angle  which  N  makes  with  Jf,  so  that  if 


tan  C 


A2B1  -  AiB2 


we  must  set 

Hence 
and 


A1A2  +  B1B2 
Ai  =  6,  Bi  =  i; 
A2  =  Q,  B2=-l. 

24  +  6  _  30  _  15 
~32~ 


tanC  = 


-4 

C  =  tan-i(if). 


16 


We  may  verify  these  results.     For  if  J5  =  -,  then  A  =  --C;  and  hence 


(6,  p.  20,  and  1,  p.  19)  tan  ^  =  cot  C 
true  for  the  values  found. 


tan  C 


,  which  is 


Ex.  2.     Find  the  equation   of  the  line   through 

(3,  5)  which  makes  an  angle  of  —  with  the  line 
X  -  y  +  6  =  0.  ^ 

•     Solution.    Let  mi  be  the  slope  of  the  required  line. 
Then  its  equation  is  (Theorem  V,  p.  95) 

(1)  y-5  =  mi(x-3). 


YJ, 

k 

4 

/ 

3 

/ 

~~. 

^\ 

^^'1 

5) 

\ 

~~ 

\ 

\ 

\  1 

\ 

0 

\ 

\ 

112  ANALYTIC  GEOMETRY 

The  slope  of  the  given  line  is  m^  =  1,  and  since  the  angle  which  (1)  makes 

with  the  given  line  is  — ,  we  have  (by  the  Corollary),  ^     \ 

3  '  ^  ^       ' 

tan^  =  :^^^ 1,  d'^Xl'^- 


3       1  +  nil  /" 

1  +  mi        ^ 

whence  mi  = ^  =  —  (2  +  V3). 

1  -  V3 

Substituting  in  (1),  we  obtain 

y-6  =  -{2+V3){x-S),_ 

or  (2  +  Vi)  X  4-  y  -  (11  +  3  V3)  =  0. 

In  Plane  Geometry  there  would  be  two  solutions  of  this  problem,  —  the 
line  just  obtained  and  the  dotted  line  of  the  figure.  Why  must  the  latter 
be  excluded  here? 

PROBLEMS 

1.  Find  the  angle  which  the  line  3x—y-\-2  =  0  makes  with  2x  +  7/  — 2=0; 
also  the  angle  which  the  second  line  makes  with  the  first,  and  show  that 
these  angles  are  supplementary.  .         3  tt    tt 

4  '  4 

2,  Find  the  angle  which  the  line 

*  (a)  2 X  —  5 ?/  +  1  =  0  makes  with  the  line  x  —  2y  +  S  =  0. 
•  (b)  X  +  y  +  1  =  0  makes  with  the  line  x  —  y  +  1  —  0. 

(c)  3x  —  4?/  +  2=0  makes  with  the  line  x  -\-  3y  —  7  =  0. 

(d)  6x  —  Sy  +  S  =  0  makes  with  the  line  x  =  6, 

(e)  x  —  'ly  +  l  =  0  makes  with  the  line  x  +  22/  —  4  =  0. 

In  each  case  plot  the  lines  and  mark  the  angle  found  by  a  small  arc. 
Ans.  (a)tan-i(-J,);  (b)|;  (c)  tan-i(V-) ;  (d)  tan-i(- i) ;  (e)  tan-^/,). 

^Z.  Find  the   angles   of  the   triangle   whose    sides    are   x  +  3y  — 4  =  0, 
3ic  -  2?/  +  1  =  0,  and  a:  -  y  +  3  =  0.  Ans.  tan-i(-  V),  tan-i(i),  tan-i(2). 

Hint.  Plot  the  triangle  to  see  wliicli  angles  formed  by  the  given  lines  are  the  angles 
of  the  triangle. 

-^4.  Find  the  exterior  angles  of  the  triangle  formed  by  the  lines  5x  —  y  +  S=0, 
y  =  2,x-4y  +  3  =  0.  Ans.    tan-i(5),  tan-i(-  i),  tan-i(- V)- 

5.  Find  one  exterior  angle  and  the  two  opposite  interior  angles  of  the 
triangle  formed  by  the  lines  2x-3?/-6=0,  3x+42/-12  =  0,  x-Sy+6=0. 
Verify  the  results  by  formula  12,  p.  20. 


THE  STRAIGHT  LINE  113 

6.  Find  the  angles  of  the  triangle  formed  by3x+2?/-4=0,  x-3y+6=0, 
and  4x  — 3y— 10=0.     Verify  the  results  by  the  formula 

tan  J.  +  tan  B  +  tan  C  =  tan  J.  tan  5  tan  C,   if  A  -\-  B -\-  C  =  180°. 

7.  Find  the  line  passing  through  the  given  point  and  making  the  given 
angle  with  the  given  line. 

.    (a)  (2,  1), -,  2x-3?/  +  2  =  0.  Ans.  5x  -  ij  -  9  =  0. 

(b)  (1,  -  3),  — ,  X  +  22/  +  4  =  0.      Ans.  3x  +  y  =  0. 

(c)  (2,  -  5),  -,  X  +  3 ?/  -  8  =  0.        Ans.  x-2ij  -12  =  0. 

/■IK    ,  V  7.  A  ?n  +  tan0   , 

(d)  (xi,  2/i),  (f>,  y  =  mx  +  b.  Ans.  y  -  yi  =  - (x  -  Xi). 

1  —  m  tan  <p 

(e)  (Xi,  2/i),  (f>,  Ax  -\-  By  -^  C  =  0.       Ans.  y  -y^  =  -—^ (x  -  Xi). 

A.  tan  <f>  -^  B 

8.  Show  from  a  figure  that  it  is  impossible  to  draw  a  line  through  the  inter- 
section of  two  lines  and  "making  equal  angles  with  those  lines"  in  the 
sense  in  which  we  have  defined  "  the  angle  which  one  line  makes  with  a 
second  line."  Prove  the  same  thing  by  formula  (X).  How  are  the  bisectors 
of  the  angles  of  two  lines  to  be  defined  ? 

9.  Given  two  lines  Xi:3x  —  4y  —  3  =  0  and  X2:4x  —  3?/  +  12  =  0;  find 
the  equation  of  the  line  passing  through  their  point  of  intersection  such  that 
the  angle  it  makes  with  Li  is  equal  to  the  angle  L^  makes  with  it. 

Ans.    7x-7?/  +  9  =  0. 

51.  Systems  of  straight  lines.  An  equation  of  the  first  degree 
in  X  and  ij  which  contains  a  single  arbitrary  constant  will  repre- 
sent an  infinite  number  of  lines,  for  the  locus  of  the  equation 
will  be  a  straight  line  for  any  value  of  the  constant,  and  the  locus 
will  be  different  for  different  values  of  the  constant. 

The  lines  represented  by  an  equation  of  the  first  degree  which 
contains  an  arbitrary  constant  are  said  to  form  a  system.  An 
equation  which  represents  all  of  the  lines  satisfying  a  single  con- 
dition must  contain  an  arbitrary  constant,  for  there  is  an  infinite 
number  of  lines  satisfying  a  single  condition  ;  hence  a  single  geo- 
metrical condition  defines  a  system  of  lines. 

Thus  the  equation  y=2x-\-b,  where  b  is  an  arbitrary  constant,  represents  the 
system  of  lines  having  the  slope  2;  and  the  equation  y  —  5  —  m  (x  —  3),  where  m 
is  an  arbitrary  constant,  represents  the  system  of  lines  passing  through  (3,  5). 


114 


ANALYTIC  GEOMETRY 


Second  rule  to  find  the  equation  of  a  straight  line  satisfying  two 
conditions. 

First  step.  Write  the  equation  of  the  system  of  lines  satisfying 
one  condition. 

Second  step.  Determine  the  arbitrary  constant  in  the  equation 
found  in  the  first  step  so  that  the  other  condition  is  satisfied. 

Third  step.  Substitute  the  result  of  the  second  step  in  the  result 
of  the  first  step.    This  gives  the  required  equation. 

This  rule  is,  in  general,  easier  of  application  than  the  rule  on 
p.  93.  It  has  already  been  applied  in  solving  Ex.  2,  p.  Ill,  and 
will  find  constant  application  in  the  following  sections.  The 
number  of  lines  satisfying  the  conditions  imposed  will  be  the 
number  of  real  values  of  the  arbitrary  constant  obtained  in 
the  second  step. 

Ex.  1.  Find  the  equations  of  the  straight  lines  having  the  slope  f  and 
intersecting  the  circle  «2  +  ^/^  =  4  in  but  one  point. 

Solution.    First  step.    The  equation 

represents  the  system  of  lines  whose  slopes  are  |  (Theorem  I,  p.  58). 

Second  step.     The  coordinates  of  the  inter- 
section of  the  line  and  circle  are  found  by  solv- 
ing their  equations  simultaneously  (Rule,  p.  76). 
\yi  ^111        Substituting  the  value  of  y  in  the  line  in  the 
V^ \. equation  of  the  circle,  we  have 

x2-f  (|«  +  &)2  =  4, 
or  25x2  4.  24 6x  +  (16 &2  _  64)  =  0. 


^: 


/ 


YA 


The  roots  of  this  equation,  by  hypothesis, 
must  be  equal;  hence  the  discriminant  must 
vanish  (Theorem  II,  p.  3) ;   that  is, 

576  &2_  100  (16  62- 64)  =  0, 


whence 


Third  step.    Substitute  these  values  of  b  in  the  equation  of  the  first  step. 
We  thus  obtain  the  two  solutions 

and  2/  =  f  X  -  f . 


THE  STRAIGHT  LINE  115 

PROBLEMS 

•    1.  Write  the  equations  of  the  systems  of  lines  defined  by  the  following 
conditions. 

(a)  Passing  through  (—2,  3). 

(b)  Having  the  slope  —  f . 

(c)  Distance  from  the  origin  is  3. 

(d)  Having  the  intercept  on  the  F-axis  =  —  3. 

(e)  Passing  through  (6,  —  1). 

(f )  Having  the  intercept  on  the  JT-axis  =  6. 

(g)  Having  the  slope  i. 

(h)  Having  the  intercept  on  the  F-axis  =  5. 
(i)  Distance  from  the  origin  =  4. 

2.  What  geometric  conditions  define  the  systems  of  lines  represented  by 
the  following  equations  ? 

(a)  2»-3y  +  4A:  =  0. 

(b)  kx-Sy  -7  =  0. 

(c)  X  +  y  -  k  =  0. 

(d)  x  +  k  =  0. 

(e)  x  +  2ky  -S  =  0. 

(f )  2kx-3y  +  2  =  0. 

(g)  X  cos  a  +  2/  sin  a  +  5  =  0. 

Hint.    Reduce  the  given  equation  to  one  of  the  well-known  forms  of  the  equation  of 
the  first  degree. 

3.  Determine  k  so  that 

(a)  the  line  2x  —  Sy  +  k  =  0  passes  through  (—2,  1).      Ans.  k  =  7. 

(b)  the  line  2kx  —  5y  +  S  =  0  has  the  slope  3.  Ans.  k  =  J/. 

(c)  the  line  x-\-  y  —  k  =  0  passes  through  (3,  4).  Ans.  k  =  7. 

(d)  the  line  Sx  —  4y  -\-  k  =  0  has  intercept  on  X-axis  =  2. 

Ans.    k  =—  6. 

(e)  the  line  x  —  Sky  +  4:  =  Q  has  intercept  on  F-axis  =  —  3. 

Ans.    k  =  —  ^. 

(f)  the  line  4x  —  3y-f6A:  =  0is  distant  three  units  from  the  origin. 

Ans.    k  =  ±  ^. 

4.  Find  the  equations  of  the  straight  lines  with  the  slope  —  j\  which  cut 
the  circle  x^  -^  y"^  =  1  in  but  one  point.  Ans.    5  x  -f- 12  y  =  ±  13. 

5.  Find  the  equations  of  the  lines  passing  through  the  point  (1,  2)  which 
cut  the  circle  x^  -{-y^  =  4:in  but  one  point.     Ans.    y  =  2  and  4  x  -|-  3  ?/  =  10. 

6.  Find  the  equation  of  the  straight  line  passing  through  (—2,  5)  which 
makes  an  angle  of  45°  with  the  F-axis.  Ans.  x  +  y  —  3  =  0. 


116 


ANALYTIC  GEOMETRY 


7.  Find  the  equation  of  the  straight  line  which  passes  through  the  point 
(2,  —  1)  and  which  is  at  a  distance  of  two  units  from  the  origin. 

Ans.   X  =  2  and  Sx  —  4y  =  10. 

8.  Find  the  equation  of  the  straight  line  whose  slope  is  f  such  that  the 
distance  from  the  line  to  the  point  (2,  4)  is  2.  Ans.    3  x  —  4  y  =  0. 

52.  The  system  of  lines  parallel  to  a  given  line. 

Theorem  XI.    The  system  of  lines  parallel  to  a  given  line 

Ax  +  By  +  C  =  0 
is  represented  by 

(XI)  Aic-\-  By  -\-k  =  Oj 

where  k  is  an  arbitrary  constant. 

Proof.  All  of  the  lines  of  the  system  represented  by  (XI)  are 
parallel  to  the  given  line  (Corollary  II,  p.  87).  It  remains  to  be 
shown  that  all  lines  parallel  to  the  given  line  are  represented  by 
(XI).  Any  line  parallel  to  the  given  line  is  determined  by  some 
point  Pi  (xi,  ?/i)  through  which  it  passes.  If  Pi  lies  on  (XI), 
then  AX]_  +  By^^  -f  A:  =  0; 

and  hence  k  —  —  Ax^  —  By^. 

That  is,  the  value  of  k  may  be  chosen  so  that  the  locus  of  (XI) 
passes  through  any  point  Pi.  Then  (XI)  represents  all  lines 
parallel  to  the  given  line.  q.e.d. 

It  should  be  noticed  that  the  coefficients  of  x  and  y  in  (XI) 
are  the  same  as  those  of  the  given  equation. 

Ex.  1.    Find  the  equation  of  the  line  through  the  point  Pi  (3,  —  2)  paral- 
lel to  the  line  Li:2x -^y  -  A  =  0. 

Solution.    Apply  the  Rule,  p.  114. 
First  step.    The  system  of  lines  parallel  to 
the  given  line  is 

2x-Sy  -\-k  =  0. 

Second    step.     The    required    line    passes 
through  Pi;   hence 

•     2-3-3(-2)  +  fc  =  0, 
and  therefore         k  =—12. 

Third  step.     Substituting  this  value  of  k, 
the  required  equation  is 

2x-3y  -12  =  0 


THE  STRAIGHT  LINE 


117 


53.  The  system  of  lines  perpendicular  to  a  given  line. 
Theorem  XII.      The  system  of  lines  perpendicular  to  the  given 

is  represented  by 

(XII)  BQe-Ay  +  k  =  0, 

where  h  is  an  arbitrary  constant. 

Proof.  All  of  the  lines  of  the  system  represented  by  (XII) 
are  perpendicular  to  the  given  line,  for  (Corollary  III,  p.  87) 
AB  —  BA  =  0.  It  remains  to  be  shown  that  all  lines  perpen- 
dicular to  the  giv.en  line  are  represented  by  (XII).  Any  line 
perpendicular  to  the  given  line  is  determined  by  some  point 
^i(^ij  l/i)  through  which  it  passes.      If  Pi  lies  on  (XII),  then 


whence 


Bxi  —  Ayi  -{-  k  =  0, 
k  =  Ayi  —  Bxi. 


That  is,  the  value  of  k  may  be  chosen  so  that  the  locus  of  (XII) 
passes  through  any  point  Pi.  Then  (XII)  represents  all  lines 
perpendicular  to  the  given  line.  q.e.d. 

Notice  that  the  coefficients  of  x  and  y  in  (XII)  are  respectively 
the  coefficients  of  y  and  x  in  the  given  equation  with  the  sign  of 
one  of  them  changed. 

Ex.  1.  Find  the  equation  of  the  line  through  the  point  Pi  (—  1,  3)  perpen- 
dicular to  the  line  Li:  6x  —  2y  ■}■  S  =  0. 

Solution.    Apply  the  Rule,  p.  114. 

First  step.    The  equation  of  the  system  of  lines  perpendicular  to  the  given 
line  is 

2x-\-  5y  +  k  =  0. 

Second    step.      The    required   line    passes 

-  —^ — I — I — r**kJ     I — [—     through  Pi ;    hence 

"/^l     M     I     I     Ml^  2(-l)  +  6-3  +  A:  =  0, 

or  A;  =  -  13. 

Third  step.    Substitute  this  value  of  k.     The  required  equation  is  then 
2x  +  62/-13  =  0. 


118  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Find  the  equation  of  the  straight  line  which  passes  through  the  point 

(a)  (0,  0)  and  is  parallel  to  (c  —  3 y  +  4  =  0.  Ans.  x  —  Sy  =  0. 

(b)  (3,  —  2)  and  is  parallel  to  x  -\-  y  -\-  2  =  0.  Ans.  x  +  y  —  1  =  0. 

(c)  (—  5,  6)  and  is  parallel  to  2x  + 'ky  -  S  =  0.     Ans.  x  +  2y  -7  =  0. 

(d)  (—1,  2)  and  is  perpendicular  toSx  —  iy  +  l  =  0. 

Ans.    Ax  +  Sy  -2  =  0. 

(e)  (—  7,  2)  and  is  perpendicular  tox-3?/  +  4  =  0. 

Ans.    Sx  +  y  +  19  =  0. 

2.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  whose  vertices  are  (-  3,  2),  (3,  -  2),  and  (0,  —  1),  which  are  parallel 
to  the  opposite  sides. 

Ans.  The  sides  of  the  triangle  are 

2x  +  3y  =  0,x-{-Sy  +  S  =  0,x-j-y-}-l  =  0. 
The  required  equations  are 

2x  +  Sy  +  S  =  0,x  +  Sy-S  =  0,x  +  y-l  =  0. 

3.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  in  problem  2  which  are  perpendicular  to  the  opposite  sides,  and 
show  that  they  meet  in  a  point. 

Ans.   Sx-2y  -2  =  0,  3x-y  +  11  =  0,  x-y- 6  =  0. 

4.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of  the 
triangle  in  problem  2,  and  show  that  they  meet  in  a  point. 

Ans.  Sx-2y  =  0,3x-y-6  =  0,x-y  +  2  =  0. 

6 .  The  equations  of  two  sides  of  a  parallelogram  are  3a;  —  4y+6  =  0  and 
X  +  5  2/  —  10  =  0.  Find  the  equations  of  the  other  two  sides  if  one  vertex 
is  the  point  (4,  9).  Ans.    Sx  —  4y  +  24:  =  0  and  x  +  5 ?/  —  49  =  0. 

6.  The  vertices  of  a  triangle  are  (2,  1),  (—  2,  3),  and  (4,  —  1).  Find  the 
equations  of  (a)  the  sides  of  the  triangle,  (b)  the  perpendicular  bisectors  of 
the  sides,  and  (c)  the  lines  drawn  through  the  vertices  perpendicular  to  the 
opposite  sides.  Check  the  results  by  showing  that  the  lines  in  (b)  and  (c) 
meet  in  a  point. 

7.  Show  that  the  perpendicular  bisectors  of  the  sides  of  any  triangle  meet 
in  a  point. 

8.  Show  that  the  lines  drawn  through  the  vertices  of  a  triangle  perpen- 
dicular to  the  opposite  sides  meet  in  a  point. 

9.  Find  the  value  of  C  in  terms  of  A  and  B  if  Ax  +  By  +  C  =  0  passes 
through  a  given  point  Pi  (xi,  yi) ;  show  that  the  equation  of  the  system  of 
lines  through  Pi  may  be  written  A{x  —  Xi)  +  B{y  —  yi)  =  0. 


THE  STRAIGHT  LINE  119 

54.  The  system  of  lines  passing  through  the  intersection  of 
two  given  lines. 

Theorem  XIII.  The  system  of  lines  passing  through  the  intersec- 
tion of  two  given  lines 

and  ig :  ^2^  +  ^2^/  +  C'g  =  0 

is  represented  by  the  equation 

(XIII)       Aix  +  Biy  +  Ci  +  ^  (^2£c  +  Bty  +  C2)  =  O, 

where  k  is  an  arbitrary  constant. 

Proof  All  of  the  lines  represented  by  (XIII)  pass  through  the 
intersection  of  L^  and  L^.  For  let  P^  (x^,  yi)  be  the  intersection 
of  Li  and  Xg.     Then  (Corollary,  p.  53) 

A^x,-\-B,y^^C^  =  0 
and  A  2X1  +  B^yi  +  C2  =  0. 

Multiply  the  second  equation  by  k  and  add  to  the  first.  This 
gives 

^1^1  +  B,y,  +  Ci  +  A^  (^23^1  +  B0,  +  C2)  =  0. 

But  this  is  the  condition  that  Pj  lies  on  (XIII). 
That  all  lines  through  the  intersection  of  Xi  and  L^  are  repre- 
sented by  (XIII)  follows  as  in  the  proofs  of  Theorems  XI  and 

XII.  Q.E.D. 

Corollary.  If  L^  and  L^  are  parallel,  then  (XIII)  represents  the 
system  of  lines  parallel  to  Li  and  L^. 


For  if  Li 

and  L2  are 

parallel,  then 

Ax_Bx 
A2      B2 

and  hence 

Ai  _Bi 
kA^      kBi 

By  composition, 

Ai  +  kA2_Bi  +  kB2 
Ai                 Bx 

Hence  L 

and  (XIII) 

are  parallel  (Corollary  II,  p.  87) 

Notice  that  (XIII)  is  formed  by  multiplying  the  equation  of  L^ 
by  k  and  adding  it  to  the  equation  of  L^. 


120 


ANALYTIC  GEOMETRY 


Ex.  1.     Find  the  equation  of  the  line  passing  through  Pi  (2,  1)  and  the 
intersection  of  ii :  3  ic  —  5  ?/  —  10  =  0  and  L2  :  x  ■{-  y  +  1  =  0. 

Solution.    Apply  the  Rule,  p.  114.    The  system  of  lines  passing  through  the 
intersection  of  the  given  lines  is  represented  by 

Sx  -  5 y  -  10  -^  k {x  +  y  -\- 1)  =  0. 
If  Pi  lies  on  this  line,  then 

6_5-10  +  fc(2  +  l  +  l)  =  0; 
whence  A;  =  |. 

Substituting  this  value  of  k  and  simplifying, 
we  have  the  required  equation 

21x-ll?/-31  =  0. 


Ex.  2.  Find  the  equation  of  the  line  passing  through  the  intersec- 
tion of  Li:2x  +  y  +  1  =0  and  L-zix  -  2y  +  1  =  0  and  parallel  to 
Ls-Ax-Sy-1  =  0. 

Solution.  Apply  the  Rule,  p.  114.  The  equation  of  every  line  through 
the  intersection  of  the  first  two  given  lines  has  the  form 

2x  +  ?/  +  l  +  A;(x-2?/  +  l)  =  0, 
or     (2  +  A:)x  +  (1  -  2 /c) ?/  +  (1  +  A:)  =  0. 

If  this  line  is  parallel  to  the  third  line  (Corollary 
II,  P-  87), 

2+fc_l-2fc 
4     ~     -3    ' 
whence  k  =  2. 

Substituting  and  simplifying,  we  obtain 
4x-3y  +  3  =  0. 

The  geometrical  significance  of  the  value  of  k  in  Theorem  XIII 
is  given  most  simply  when  L^  and  L2  are  in  normal  form. 

Theorem  XIV.    The  ratio  of  the  distances  from 
Li :  X  cos  0)1  +  2/  sin  toi  —  ^1  =  0 
and  L2 :  x  cos  (1)2  -{-  y  sin  (02  —  2^2  =  ^ 

to  any  'point  of  the  line 

L  :  X  cos  (01  +  2/  sin  wi  —  pi  -\-  k  (x  cos  wo  +  ?/  sin  wg  —  ^2)  =  0 
is  constant  and  equal  to  —  k. 


THE  STRAIGHT  LINE  121 

Proof.    Let  Pi  {x^  ?/i)  be  any  point  on  L.     Then 

Xi  cos  0)1  +  yx  sin  ini—  pi-\-  k  (x^  cos  <i>2  +  yi  sin  wg  —  p^  =  0, 

and  hence  -  h  =  "'i  cos.». +  y.  sin.», -y, 

Xi  cos  0)2  +  2/i  sm  0)2  —  ^2 

The  numerator  of  this  fraction  is  the  distance  from  Zi  to  Pi, 
and  the  denominator  is  the  distance  from  L^  to  Pi  (Theorem  IX, 
p.  106).  Hence  —  A;  is  the  ratio  of  the  distances  from  L^  and  L^ 
to  any  point  on  L.  q.e.d. 

Corollary.  If  k  =  ±  1,  then  L  is  the  bisector  of  one  of  the  angles 
formed  by  L^  and  L^.  That  is,  the  equations  of  the  bisectors  of  the 
angles  between  tivo  lines  are  found  by  reducing  their  equations  to 
the  normal  form  and  adding  and  subtracting  them. 

For  when  Jfc  =  ±  1  the  numerical  values  of  the  distances  from  L\  and  L2  to  any 
point  of  L  are  equal. 

The  angle  formed  by  L^  and  L^  in  which  the  origin  lies,  or  its 
vertical  angle,  is  called  an  internal  angle  of  L^  and  igj  ^^^  either 
of  the  other  angles  formed  by  Xi  and  L^  is 
called  an  external  angle  of  those  lines.  From 
the  rule  giving  the  sign  of  the  distance  from  a 
line  to  a  point  (p.  105)  it  follows  that  L  lies  in 
the  internal  angles  of  L^  and  L^  when  k  is  nega- 
tive, and  in  the  external  angles  when  k  is  posi- 
tive. If  the  origin  lies  on  L^  or  L^,  the  lines 
must  in  each  case  be  plotted  and  the  angles  in  which  k  is  posi- 
tive found  from  the  figure. 


PROBLEMS 

1.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
2x-3?/  +  2  =  0  and  3x-4?/— 2  =  0,  ^ithout  finding  the  point  of  intersec- 
tion,) which 

(a)  passes  through  the  origin. 

(b)  is  parallel  to5x  —  2y  +  3  =  0. 

(c)  is  perpendicular  to3x  —  2?/  +  4  =  0. 

Ans.  {B.)bz-ly  =  0;  (b)  5x  -  2^/  -50  =  0  ;  (c)  2x  +  By  -  58  =  0. 


122  ANALYTIC  GEOMETRY 

2.  Find  the  equations  of  the  lines  which  pass  through  the  vertices  of  the 
triangle  formed  by  the  lines  2x  —  Sy+l  =  0,  x  —  y  =  0,  and  3x  +  4?/  —  2  =  0 
which  are 

(a)  parallel  to  the  opposite  sides. 

(b)  perpendicular  to  the  opposite  sides. 

Ans.    (a)  3x  +  4  2/- 7  =  0,  14x-21?/+    2  =  0,  17x  -  ITy  +  5  =  0; 
(b)  4:X-Sy-1=0,  21x  +  Uy  -10  =  0,  17x  +  17?/ -  9  =  0. 

3.  Find  the  bisectors  of  the  angles  formed  by  the  lines  Ax  —  3y  —  1  =  0 
and  3x  —  4^  +  2  =  0,  and  show  that  they  are  perpendicular. 

Ans.    7  X  -7y  -\-  1  =  0  and  x  +  ?/  —  3  =  0. 

4.  Find  the  equations  of  the  bisectors  of  the  angles  formed  by  the  lines 
6x-12  2/  +  10  =  0  and  12x  -  5?/  + 15  =  0.    Verify  the  results  by  Theorem  X. 

5.  Find  the  locus  of  a  point  the  ratio  of  whose  distances  from  the  lines 
4x-3i/  +  4  =  0and5x+12y-8=:0isl3to5.       Ans.    9x  +  9y-i  =  0. 

6.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by  the 
lines  4x-3y  =  12,  5x-12i/-4  =  0,  and  12x-5y-13  =  0.  Show  that 
they  meet  in  a  point. 

Ans.    7x-9y-16  =  0,  7x  +  7y-9  =  0,  112x-642/-221=0. 

7.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by  the 
lines  5x-12y  =  0,  5 x  +  12 y  +  60  =  0,  and  12 x  -  5 ^  -  60  =  0,  and  show 
that  they  meet  in  a  point. 

Ans.    2y  +  6  =  0,  17 x  +  7 ?/  =  0,  17 x  -  17 y  -  60  =  0. 

8.  The  sides  of  a  triangle  are  3x  +  4?/  —  12  =  0,  Sx  —  iy  =  0,  and 
4x  +  3?/  +  24  =  0.  Show  that  the  bisector  of  the  interior  angle  at  the 
vertex  formed  by  the  first  two  lines  and  the  bisectors  of  the  exterior  angles 
at  the  other  vertices  meet  in  a  point. 

9.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
x  +  y  —  2  =  0  and  x  —  y  +  Q  =  0  and  through  the  intersection  of2x  —  y  +  3  =  0 
andx -3?/  + 2  =  0.  Ans.    19x -{- Sy  +  26  =  0. 

Hint.  The  systems  of  lines  passing  through  the  points  of  intersection  of  the  two  pairs 
of  lines  are 

x  +  y  —  2+k{x-y  +  G)  =  0 

and  2x  —  y  +  3  +  k'(x  —  3y  +  2)=0. 

These  lines  will  coincide  if  (Theorem  III,  p.  88) 

l  +  k_     1-k     _-2  +  6Jc 
2+k'~-l-3k'      3+2k'  ' 

Letting  p  be  the  common  value  of  these  ratios,  we  obtain 

l  +  k=2p  +  pk% 
l-k  =  -p-3pk', 
and  -2  +  6fc=3p  +  2p/y. 

From  these  equations  we  can  eliminate  the  terms  in  pk'  and  p,  and  thus  find  the  value 
of  k  which  gives  that  line  of  the  first  system  which  also  belongs  to  the  second  system. 


THE  STRAIGHT  LINE  f23 

10.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
2X  +  52/  —  3  =  0  and  Sx  -  2y  —  1  =  0  and  through  the  intersection  of 
x-y  =  0  and  x  +  3y-6  =  0.  Ans.   43  x  -  35  y  -  12  =  0. 

A  figure  composed  of  four  lines  intersecting  in  six  points  is 
called  a  complete  quadrilateral.  The  six  vertices  determine  three 
diagonals  of  which  two  are  the  diagonals  of  the  ordinary  quadri- 
lateral formed  by  the  four  lines. 

1 1 .  Find  the  equations  of  the  three  diagonals  of  the  complete  quadrilateral 
formed  by  the  lines  x  -{■  2y  =  0,  3x-4?/  +  2  =  0,  x  -  y  -\-  S  =  0,  and 
3x-2?/  +  4  =  0.        Ans.   2x-?/  +  l  =  0,  a;  +  2  =  0,  5x-6y  +  8  =  0. 

12.  Show  that  the  bisectors  of  the  angles  of  any  two  lines  are  perpen- 
dicular. 

13.  Find  a  geometricarinterpretation  of  k  in  (XI)  and  (XII). 

14.  Find  the  geometrical  interpretation  of  ^  in  (XIII)  when  Li  and  L^ 
are  not  in  normal  form. 

15.  Show  that  the  bisectors  of  the  interior  angles  of  any  triangle  meet  in 
a  point. 

16.  Show  that  the  bisectors  of  two  exterior  angles  of  a  triangle  and  of  the 
third  interior  angle  meet  in  a  point. 

55.  The  parametric  equations  of  the   straight  line.     The 

angles  a  and  ^  between  a  line  directed  upward  "*  and  the  coordi- 
nate axes  (p.  28)  are  called  the  direction  angles  of  the  line. 
Their  cosines,  cos  a  and  cos  fi,  are  called  the  direction  cosines  of 
the  line  and  satisfy  the  relation 

(1)  cos^a-f  cos2^  =  1. 

For  (Theorem  I,  p.  28)  cos  ^  =  sin  or  and  sin^  a  +  cos^  a  =  1. 

Given  a  line  with  direction  angles  a  and  /3  passing  through 
Pi(xi,  7/i).  Let  P(x,  y)  be  any  point  on  this  line  and  denote  the 
variable  directed  length  P^P  by  p.  The  projections  of  PyP  on 
the  axes  are  respectively  (Theorem  III,  p.  31) 

ic  —  iCi  and  y  —  ?/i, 
or  (Theorem  II,  p.  30) 

p  cos  a  and  p  cos  ^. 

*  If  the  line  is  horizontal  we  suppose  that  it  is  directed  to.  the  right,  so  a=  0  and  i3  =  -• 


124  ANALYTIC  GEOMETRY 

Hence        x  —  Xi  =  p  cos  a  and  y  —  y^  =  p  cos  ^; 
whence  x  =  x-^-{-  p  cos  a. 

y  =  yi-\-p  (^os  /3. 
Hence  we  have 

Theorem  XV.  Parametric  form.  The  coordinates  of  any  pomt 
P  (x,  y)  on  the  line  through  a  given  point  Pi  (xi,  y^  whose  direction 
angles  are  a  and  /3  are  given  by 

^      ^  \y  =  yi  +  pcosfi, 

where  p  denotes  the  variable  directed  length  PiP. 

Equations  (XV)  are  called  the  parametric  equations  of  the 
straight  line  because  they  express  the  variable  coordinates  of  any 
point  (x,  y)  on  the  line  in  terms  of  a  single  variable  parameter  p. 
As  p  varies  from  —  oo  to  +  oo  the  point  P  {x,  y)  describes  the  line 
in  the  positive  direction.  These  equations  are  important  in  deal- 
ing with  problems  which  involve  the  distances  from  a  point  7\ 
on  a  line  to  the  intersections  of  that  line  with  a  given  curve. 

Theorem  XVI.  Symmetric  form.  The  equation  of  a  straight  line 
in  terms  of  the  coordinates  of  a  point  Pj  (xi,  y^  on  the  line  and 
its  direction  cosines  is 

^         ^  cos  a    ""    cos  p 

Hint.   Solve  (XV)  for  p  and  equate  the  two  values  obtained. 

Theorem  XVII.    The  direction  cosines  of  the  line 
Ax  -^  By  -[-C  =  0 

-^  o 

are  cos  o  = .   cos  p 


when  the  sign  of  the  radical  is  the  same  as  that  of  A . 

Proof.    Let  Pi  {x^.  y^  be  a  point  on  the  given  line.     Then 
(Corollary,  p.  53)  ^^^  ^  ^^^  +  (7  =  0. 


THE  STRAIGHT  LINE  125 

SubtractiDg  from  the  given  equation,  we  obtain 

A(x-x{)-hB(t/-y,)=0. 

Transpose  the  second  term  and  divide  by  —AB]  this  gives 

a;  —  a^i  ^  y  —  yi 
-B  A 

Dividing  this  equation  by  (XVI),  we  have 

cos  a  _  cos  /3 
-B  ~     A    ' 

Let  r  denote  the  common  value  of  these  ratios.     Then 

cos  a  =  —  Br  and  cos  (3  =  Ar. 

Squaring  and  adding, 

cos^  a  +  cos2  13  =  (.1^  4-  B^)  r\ 

Then  from  (1),  p.  123,  r  = y 

and  hence 

—  B  A 

(2)  cos  a  — --==^=  and  cos  (3 


The  sign  of  the  radical  must  be  the  same  as  that  of  yl.        q.e.d. 
[For  since  the  line  is  directed  upward,  i3<  — ,  and  hence  cos/3  is  positive.] 

Corollary.  If  cos  cc  and  cos  f3  are  proportional  to  two  numbers 
a  and  h,  then 

a  o  ^ 

cos  a  = ?   cos  p  = 

±  Va^  +  ft2  +  Va^  +  ft2 

The  sign  of  the  radical  must  be  the  same  as  that  of  b. 

To  reduce  the  equation  of  a  given  straight  line  to  the  symmet- 
rical or  parametric  form  it  is  necessary  to  know  the  coordinates 
of  some  point  on  the  line  (which  may  be  found  by  the  Rule, 
p.  60)  and  its  direction  cosines  (which  are  given  by  Theorem 
XVII).  Then  we  can  write  the  required  equations  by  Theorem 
XV  or  XVI. 


126 


ANALYTIC  GEOMETRY 


P  X   '      y 

0  2.      1 

5-1       6 


Ex.  1.    Plot  the   line  whose  parametric  equations  are  x  =  2  —  |  /o  and 

Solution.  Comparing  with  (XV)  we  see  that 
Pi  (2,  1)  is  a  point  on  the  line.  A  second  point 
will  enable  us  to  plot  the  line.  We  have  at 
once  the  table 

Hence  the  line  joining 
the  points  Pi  (2,  1)  and 
P2(—  1,  5)  is  the  required 
line. 

(x,2/),  or(2-fp,  l  +  4p), 
are  the  coordinates  of  that 
variable  point  P  on  the  line  whose  distance 
from  Pi  is  the  variable  p. 

Ex.  2.  Given  the  circle  C  :  x^  -{■  y^  =  2,6  and  the  line  whose  parametric 
equations  are  x  =  5  —  f  p  and  y  =  —  3  +  |  /? ;  find  the  product  of  the  dis- 
tances from  Pi  (5,  —  3)  to  the  points  of  intersection  of  the  .line  with  C,  and 
the  middle  point  of  the  chord  formed  by  the  line. 

Solution.    By  Theorem  XV  the  coordinates  of  any  point  on  the  line 
are    (5-fp,    -3  +  fp),    where   p 
denotes    the   distance    from    Pi    to 
that  point.      If  that  point  lies  on 
C  (Corollary,  p.   53), 

(5-3^)2  +  (_3  +  |p)2  =  25, 

or,  simplifying, 


(3) 


/)  +  9  =  0. 


The  roots  of  this  quadratic  are 
the  directed  lengths  pi  =  P1P2  and 
pi  =  P1P3,  where  P^  and  P3  are  the 

points  of  intersection  of  the  line  and 
circle.  For  if  P2  (5  -  |  pi,  -  3  +  f  pi) 
is  on  the  circle, 

(5-|Pi)^+(-3  +  |pi)2  =  25, 


or 


Pi' 


Upi  +  9  =  0. 


Hence  pi,  and  similarly  p2,  is  a  root  of  (3). 
The  product  of  these  distances  is  therefore  9  (Theorem  I,  p.  3). 
Half  the  sum  of  these  roots  is  PiP,  or  2/-  (Theorem  I,  p.  3).     For  p  =  ^^ 
we  have  x  =  f |  and  y  =  ||,  so  the  middle  point  of  the  chord  is  the  point 


THE  STRAIGHT  LINE  127 

PROBLEMS 
1.  Plot  the  following  lines : 


('^  {:=2;S:.     .     <^» 


1 

V5 
2 

V5 


2.  Prove  that  if  cos  a  and  cos  /3  are  the  direction  cosines  of  a  line  directed 
upward,  then  —  cos  a  and  —  cos  /3  are  the  direction  cosines  of  the  same  line 
directed  downward. 

3.  Find  the  coordinates  of  the  points  on  the  line  H      ~      -.  ^  ,    for  which 

p  =  S,  —  2,  and  4.     Verify  the  geometric  significance  of  p  for  each  of  these 
points  by  Theorem  IV,  p.  31. 

4.  Find  the  product  of  the  distances  from  Pi  (2,  1)  to  the  intersections  of 
the  line  x  =  2  —  ^p  and  y  =  1  +  | p  with  the  circle  x^  -\-  y^  =  25,  and  explain 
the  sign  of  the  result.  Ans.    —  20. 

6.  Given  the  ellipse  x^  +  4:y^  =  16  and  the  line  x  =  Xi  -  ^ p  and 
y  =  Vi  +  ^  P ;  find  the  equation  whose  roots  are  the  distances  from 
Pi{^i,  Vi)  to  the  points  of  intersection  of  the  line  and  ellipse. 

Ans.   -p^  -  ^"^^  ~  ^^^'p  +  xi2  +  4yi2  -  16  :^  0. 
25  5 

6.  Find  the  condition  that  Pi  in  problem  5  should  be  the  middle  point  of 
the  chord  on  which  it  lies. 

Hint.    The  two  values  of  p  must  be  numerically  equal,  with  opposite  signs. 

7.  Given  the  parabola  y^  =  4x  and  the  line  a;  =  2  +  p cosar,  y  =  -4: 
-I-  p  cos  /3 ;  find  the  condition  which  cos  a  and  cos  j8  must  satisfy  if  the 
line  meets  the  parabola  in  but  one  point. 

Ans.   cos2  a  +  4  cos  a  cos  /3  +  2  cos^/S  =  0. 

8.  If  a  and  6  are  two  numbers  such  that  a^  +  62  =  1,  prove  that  a  and  b 
are  the  direction  cosines  of  some  line. 

9.  Derive  equation  (XVI)  from  Theorem  V  (p.  95)  and  Theorem  I  (p.  28). 

10.  Prove  that  the  common  value  of  the  ratios  in  (XVI)  is  the  length  PiP. 

Hint.  Square  (XVI),  apply  the  Theorem  on  the  sum  of  the  antecedents  and  of  the 
consequents,  and  then  take  the  square  root. 

11.  Derive  equations  (XV)  from  (XVI)  by  means  of  problem  10. 


^6 
128  ANALYTIC  GEOMETRY 

MISCELLANEOUS   PROBLEMS 

1.  Find  the  point  on  the  line  3  x  —  5  ?/  +  6  =  0  which  is  equidistant  from 
the  points  (3,  —  4)  and  (2,  1). 

2.  Find  the  equation  of  the  line  through  the  intersection  of  the  lines 
7x  +  2/-3  =  0  and  3x  +  6y  —  11  =  0  which  is  perpendicular  to  the  line 
joinhig  their  intersection  to  the  origin. 

3.  Find  the  equation  of  the  line  through  the  point  (2,  5)  such  that  the 
portion  of  the  line  included  between  the  axes  is  bisected  at  that  point. 

4.  Find  the  equation  of  the  line  through  the  point  (2,  —  3)  such  that 
the  portion  of  the  line  included  between  the  lines  3x  +  2/  —  2  =  0  and 
x  +  5?/  +  10  =  0  is  bisected  at  that  point. 

6.  Prove  that  the  diagonals  of  a  rhombus  are  perpendicular. 

6.  If  the  F-axis  makes  an  angle  of  w  with  the  X-axis,  find  the  equation  of 
the  straight  line  in  terms  of  its  intercept  h  on  the  Y-axis  and  its  inclination  a. 

7.  If  the  Y-axis  makes  an  angle  of  w  with  the  JT-axis,  find  the  equation 
of  the  straight  line  whose  inclination  is  a  which  passes  through  Pi(xi,  2/i)- 

8.  If  the  Y-axis  makes  an  angle  of  w'  with  the  X-axis,  find  the  normal 
form  of  the  equation  of  the  straight  line. 

9.  Find  the  tangent  of  the  angle  which  one  line  makes  with  another  if  the 
axes  are  oblique. 

10.  Show  that  all  of  the  lines  for  which  m  =  b  pass  through  the  same 
point,  and  find  the  coordinates  of  that  point. 

1 1 .  Show  that  all  of  the  lines  for  which  -  +  -  =  constant  pass  through  the 

a      b 

same  point,  and  find  the  coordinates  of  that  point. 

12.  Prove  that  all  of  the  lines  Ax  -\- By  +  C  =  0  for  which  A  -\-  B  +  C  =  0 
pass  through  the  same  point,  and  find  the  coordinates  of  that  point. 

13.  Find  the  points  in  which  the  lines  2x  —  Sy  =  0,  x-t-4y  —  2  =  0, 
2x-3y  +  X(x+4?/-2)  =  0,  2x-3?y-X(x-  4 ?/ -  2)  =  0  cut  the  X-axis. 
Show  that  the  last  two  points  divide  the  line  joinhig  the  first  two  points  inter- 
nally and  externally  in  the  same  numerical  ratio. 

14.  Prove  that  Ax  -\-  By  +  C  =  0  represents  a  straight  line  by  showing 
that  if  Pi  and  P2  lie  on  the  locus  of  the  equation,  the  point  which  divides  PiP2 
in  the  ratio  X  lies  on  the  locus  of  the  equation. 

15.  Find  the  bisectors  of  the  exterior  angles  of  the  triangle  formed  by 
2  X  -  3  y  -f- 120  =  0,  X  +  ?/  =  0,  and  3x-f-42/-6  =  0.  Show  that  these  lines 
meet  the  opposite  sides  in  three  points  on  the  same  straight  line. 


THE  STRAIGHT  LINE  129 

16.  Find  the  equation  of  tlie  line  passing  through  the  intersection  of 
Az  +  By  -^  C  =  0  and  A'x  +  B'y  +  C  =  0  which  (a)  passes  through  the 
origin,  (b)  is  parallel  to  the  X-axis,  (c)  is  parallel  to  the  F-axis. 

17.  Show  that  the  lines  {A  +  X^^^  +  (^  +  '^B')y  +  {C  +  XC)  =  0  pass 
through  a  point  if  X  is  a  variable  parameter  and  the  other  letters  are  constant. 

18.  Let  Aix  +  Biy  +  Ci  =  0,  A^x  +  B^y  +  Cs  =  0,  and  A^x  +  ^sy  +  C3  =  0 
be  three  given  lines  forming  a  triangle.  Show  that  the  equation  of  any  line 
Ax  -\-  By  -\-  C  =  0  may  be  written  in  the  form 

a  (Aix  +  Biy  +  d)  +  )8  {A^x  +  ^22/  +  C2)  +  7  (^3^  +  Bsy  +  C3)  =  0, 
where  a,  /3,  and  7  are  definite  constants. 
Hint.   Use  Theorem  III,  p.  88. 

19.  Find  the  ratio  in  which  the  line  2x  —  6y  +  8  =  0  divides  the  line  join- 
ing the  points  Pi(l,  3)  and  P2(7,  2). 

Hint.  The  coordinates  of  the  point  dividing  PjPa  'i^to  segments  whose  ratio  is  A  are 

— ,  ) ;  determine  A  so  that  this  point  lies  on  the  given  line. 

1  +  A.       l  +  A/' 

20.  Find  the  ratio  in  which  the  line  x  +  Sy  —  6  =  0  divides  the  line 
joining  (- 3,  2)  and  (6,  1). 

21.  Determine  m  so  that  the  line  y  =  mx  —  7  divides  the  line  joining 
(3,  2)  and  (1,  4)  in  the  ratio  3:2. 

22.  Find  the  equation  of  the  line  passing  through  the  point  (2,  —  3)  which 
divides  the  line  joining  (6,  3)  and  (2,  —  1)  in  the  ratio  2  :  5. 

23.  Show  that  the  ratio  of  the  distances  from  the  line  Ax  -}-  By  -^  C  =  0to 

the  points  Pi{Xi,  yi)  and  P2(X2,  2/2)  is  -^ -^ -• 

Ax2  +  By2  +  C 

24.  Show  that  the  line  Ax  +  By  +  C  =  0  divides  the  line  joining  Pi(xi,  yi) 
and  P2  (X2,  2/2)  into  segments  whose  ratio  is  —  - — ~ . 

25.  Show  by  the  preceding  example  that  any  line  cuts  the  sides  of  a  tri- 
angle P1P2,  P2-P3,  and  P3P1  in  the  points  L,  M,  iV^such  that 

PiL      P2M      PsN  ^     ^ 
LP2  ^  MPs  ^  NPi  ~ 

26.  Plot  the  line  2x  —  Sy  +  6  =  0  and  indicate  all  of  the  points  for  which 
2x-3y  +  5>0. 

27.  Find  the  area  of  the  triangle  formed  by  AiX  +  Biy  +  Ci  =  0, 
A2X  +  B2y  4  C2  =  0,  and  A^x  +  B^y  -f  O3  =  0. 


CHAPTER  V 
THE  CIRCLE  AND  THE  EQUATION  x^  +  y^  +  Dx  +  Ey  +  F  =  0 

56.  The  general  equation  of  the  circle.  If  (a,  ^)  is  the  center 
of  a  circle  whose  radius  is  r,  then  the  equation  of  the  circle  is 
(Theorem  II,  p.  58) 

(1)  x^-{-y^-2ax-2py-\-a^-\-  ft^-r^  =  Oj 
or 

(2)  (oc-ay  +  (y-py  =  r\   ' 

In  particular,  if  the  center  is  the  origin,  a  =  0,  ^  =  0,  and  (2) 
reduces  to 

(3)  ijc^  +  y^  =  r\ 

Equation  (1)  is  of  the  form 

(4)  x''-{-2f-^Dx-{-Ey-\-F=  0, 
where 

(5)  D  =  -2a,  E=- 2  13,  Siud  F=:a^  +  /3^- 7^. 

Can  we  infer,  conversely,  that  the  locus  of  every  equation  of 
the  form  (4)  is  a  circle  ?  By  comparing  (4)  with  (1)  we  obtain  (5). 
Whence 

(6)  a=--,  ^  =  --,  and  r^  = 

These  values  of  a  and  f3  are  real,  and  if  D^  +  E^  —  4  F  is  posi- 
tive, the  value  of  r  is  real  and  the  locus  of  (4)  is  a  circle. 

To  plot  the  locus  of  (4)  by  points  (Rule,  p.  60),  we  solve  for  y. 
This  gives 

E         I  /E^  —AF 

(7)  y  =  -'^±^~x^-Dx  +  {-^— 

The  discriminant  of  the  quadratic  under  the  radical  in  (7)  is 

©  =  D2  -  4(-  1)  f 1  =  2)2  +  ^'  -  4F, 

which  is  the  numerator  of  r^  in  (6). 

130 


THE  CIRCLE  131 

If  ©  is  positive,  the  quadratic  under  the  radical  is  positive  for 
values  of  x  between  the  roots  (Theorem  III,  p.  11)  and  the  equa- 
tion has  a  locus,  as  we  have  seen. 

If  0  is  zero,  the  roots  of  the  quadratic  are  real  and  equal  (Theo- 
rem II,  p.  3).  But  for  all  other  values  of  x  the  quadratic  is 
negative  (Theorem  III,  p.  11).     The  locus  therefore  consists  of 


the  single  point 


D         e\ 


For  the  quadratic  in  (7)  equals  zero  when  k  =  —  —  (p.  2),  and  hence,  from  (7), 

E 
the  corresponding  value  of  t/  is  —-^-     This  also  follows  from  (6)  if  we  suppose  r 

approaches  zero,  for  then  the  circle  consists  only  of  its  center  (  —  ^ »  ~  "^  ) ' 

If  0  is  negative,  the  quadratic  in  (7)  is  negative  for  all  values 
of  X  (Theorem  III,  p.  11)  except  the  roots,  which  are  imaginary 
(Theorem  II,  p.  3).     Hence  there  is  no  locus. 

The  expression  ®  =  D^  -\-  E^  —  4:F  is  called  the  discriminant 
of  (4).  When  0  =  0  the  locus  of  (4)  is  often  called  a  point-circle 
or  a  circle  whose  radius  is  zero. 

We  have  thus  proved 

Theorem  I.    The  locus  of  the  equation 

(I)  oe^  +  y^  +  Utoo  +  Ey  +  F  =  0, 

whose  discriminant  is  ®  =  D^  -{-  E^  —  4:  F,  is  determined  as  follows: 
(^a)  When  0  is  positive  the  locus  is  the  circle  whose  center  is 

(  —  —  >  —  77 )  and  whose  radius  is  r  =  -i-  Vz)^  -^  ^^  —  4i^=  ^  V©. 

^  /  (     D         e\ 

(b)  When  ©  is  zero  the  locus  is  the  point-circle  I  —  "^'  ~~  "o  7' 

(c)  When  0  is  negative  there  is  no  locus.  ^  ' 

Corollary.  When  E  =  0  the  center  of  (I)  is  on  the  X-axis,  and 
when  D  =  0  the  center  is  on  the  Y-axis. 

Whenever  in  what  follows  it  is  said  that  (I)  is  the  equation 
of  a  circle  it  is  assumed  that  0  is  positive. 


132 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  locus  of  the  equation  x^  +  y^  —  4:X  +  Sy  —  5  =  0. 

Solution.  The  given  equation  is  of 
the  form  (I),  where 

D  =  -  4,  ^  =  8,  F  =  -6, 
and  hence 

e  =  16  +  64  +  20  =  100>0. 

The  locus  is  therefore  a  circle  whose 
center  is  the  point  (2,  —  4)  and  whose 
radius  is  -J-  VlOO  =  5, 

The  equation  Ax^  -\-  Bxy  +  Cy"^ 
-{-DX  +  Ey  +  F  =0  is  called  the 
general  equation  of  the  second  degree 
in  X  and  y  because  it  contains  all 
possible  terms  in  x  and  y  of  the  second  and  lower  degrees. 

Theorem  II.    The  locus  of  the   general   equation  of  the  second 


YA 

t^ 

^ 

^ 

N 

X 

' 

/ 

0 

\ 

i-' 

/ 

\ 

\ 

(-2, 

i) 

\ 

/ 

\ 

/ 

\ 

/ 

^ 

^ 

^ 

y 

r* 

degree^ 


Ax^  +  Bxy  ^Cif^Bx^Ey^F=^, 

1)2  +  ^2 


is  a  circle  when  and  only  when  A 
is  positive. 


C,B  =  0,  and 


4.AF 


Proof  The  equation  of  every  circle  must  have  the  form  (I) ; 
hence  the  coefficients  of  x^  and  y^  must  be  equal  and  the  xy  term 
must  be  lacking ;  that  is,  the  locus  of  (II)  can  be  a  circle  only 
when  A  =C  and  B  =  0.  If  these  conditions  be  satisfied,  (II) 
may  be  written  in  the  form 

x""  +  y^  -}-  jx  -h  jy  -\-  -j  =  0, 

whose  locus  is  a  circle  when  and  only  when  its  discriminant 
D'^-^  E^-4.AF 


IS  positive. 


Q.E.D. 


57.  Circles  determined  by  three  conditions.   The  equation  of 
any  circle  may  be  written  in  either  one  of  the  forms 

(x-ay  +  (y-py  =  r' 
or  x'^-\-y^  +  Dx  +  Ey-{-  F=0. 


Cf*, 


THE  CIRCLE 


133 


Each  of  these  equations  contains  three  arbitrary  constants. 
To  determine  these  constants  three  equations  are  necessary,  and 
as  any  equation  between  the  constants  means  that  the  circle  sat- 
isfies some  geometrical  condition,  it  follows  that  a  circle  may  be 
determined  to  satisfy  three  conditions. 

Rule  to  determine  the  equation  of  a  circle  satisfying  three 
conditions. 

First  step.    Let  the  required  equation  he 

(1)  (^_a)2-f(y-/3)2  =  r2 
or 

(2)  x^-^y^  +  Dx-\-Ey^F=0, 

as  77?  ay  be  more  convenient. 

Second  step.  Find  three  equations  between  the  constants  a,  fi, 
and  r  [or  D,  E,  and  F]  which  expy^ess  that  the  circle  (1)  \_or  (2)] 
satisfies  the  three  given  co7iditions. 

Third  step.  Solve  the  equations  found  in  the  second  step  for  a,  /?, 
and  r  [or  D,  E,  and  F~\. 

Fourth  step.  Substitute  the  results  of  the  third  step  in  (1)  [or 
(2)].     The  result  is  the  required  equation. 

Ex.  1.  Find  the  equation  of  the  circle  passing  through  the  three  points 
Pi(0,  1),  P2(0,  a),  and  P3(3,  0). 

Solution.  First  step.  Let  the  required  equa- 
tion be 

(3)  x'^  +  y"^  +  Dx  +  Ey  +  F  =  0. 

Second  step.  Since  Pi,  P2,  and  Pg  lie  on  (3), 
their  coordinates  must  satisfy  (3).  Hence  we 
have 

1  +  ^  +  P  =  0, 
36  +  6  ^  -I-  P  =  0, 


(4) 
(5) 
and 

(6) 


9  +  3D  +  P  =  0. 

Third  step.    Solving  (4) ,  (5) ,  and  (6) ,  we  obtain 

E  =  -7,  F=6,  D  =  -B. 
Fourth  step.    Substituting  in  (3),  the  required  equation  is 


x2  +  y' 


7y  +  G  =  0. 


134 


ANALYTIC  GEOMETRY 


since  it 


By  Theorem  I  we  find  that  the  radius  is   |  V2  *  and  the  center  is  the 
point  (I,  I). 

Ex.  2.    Find   the    equation   of   the    circle    passing   through   the    points 
Pi  (0,  —  3)  and  P^  (4,  0)  which  has  its  center  on  the  line  ic  +  2  ?/  =  0. 

Solution.    First  step.    Let  the  required  equation  be 
(7)  x2  +  ?/2  +  Dx  +  Ey  +  F  =  0. 

Second  step.    Since  Pi  and  P2  lie  on  the  locus  of  (7),  we  have 

(8)  9-3^  +  P  =  0 
and 

(9)  16  +  4  D  +  P  =  0. 

The  center  of  (7)  is  ( ,  —  7- )  >  and 

lies  on  the  given  line, 

-f-(-f)-. 

or 

(10)  D-\-2E  =  0. 
Third  step.    Solving  (8),  (9),  and  (10),  we  obtain 

D  =  -  -V-,  E  =  h  and  P  =  -  -2/-. 

Fourth  step.    Substituting  in  (7),  Xve  obtain  the  required  equation, 

x2  +  2/2  -  -V- «  +  1  y  -  ¥  =  0. 
or  5  x2  +  5  2/2  _  14  x  +  7  ?/  -  24  =  0. 

The  center  is  the  point  (|,  —  ^^),  and  the  radius  is  ^  V29. 
> 

PROBLEMS 


1.  Find  the  equation  of  the  circle  whose  center  is 


(a)  (0,  1)  and  whose  radius  is  3. 

(b)  (—2,  0)  and  whose  radius  is  2. 

(c)  (—3,  4)  and  whose  radius  is  5. 

(e)  (or,  0)  and  whose  radius  is  a. 

(f )  (0,  p)  and  whose  radius  is  /3. 
is)  (Oj  —  jS)  and  whose  radius  is  /3. 


Arts.  x2  +  ?/2  —  2  y  -  8  =  0. 

Ans.  «2  +  2/2  ^  4  a;  _  0. 

Ans.  x^  +  y'2  -{-6x  -  Sy  =  0. 

Ans.  a;2  +  2/2  —  2  ax  =:  0. 

Ans.  x^  +  y^  -  2  Py  =  0. 

Ans.  a;2  +  2/2  +  2  ]82/  =  0. 


*  The  radius  is  easily  obtained,  since  Vi  is  the  length  of  the  diagonal  of  a  square 
whose  side  is  one  unit.  We  may  construct  a  line  whose  length  is  y/n  by  describing  a 
semicircle  on  a  line  Avhose  length  is  n  +  1  and  erecting  a  perpendicular  to  the  diameter 
one  unit  from  the  end.    The  length  of  that  perpendicular  will  be  -y/n. 


THE  CIRCLE  ^  135 

2.  Find  the  locus  of  the  following  equations. 

L  (a)  a;2  +  2/2  _  6x  -  16  =  0.  (f)  x2  +  y2  _  e^;  +  4y  -  5  =  0. 

P  — X^(b)  3x2  +  3y2_i0x-24  2/  =  0.  (g)  (x  +  1)2  +  (y  _  2)2  =  0. 

'     (c)  x2  +  2/2  ^  0.  (h)  7  x2  +  7  i/2  -  4  X  -  y  =  3. 

^    (d)  x2  +  2/2  _  8x  -  0  ?/  +  25  =  0.  (i)  x2  +  2/^  +  2  ox  +  2  6?/  +  a2  +  62  =  q. 

(e)  x2  +  2/2  -  2x  +  2 2/  +  5  =  0.  (j)  x2  +  2/2  +  16x  +  100  =  0. 

3.  Find  the  equation  of  the  circle  which 

(a)  has  the  center  (2,  3)  and  passes  through  (3,  —  2). 

Ans.    X2  +  2/2 -4X-62/-13  =  0.      '^ 

(b)  passes  through  the  points  (0,  0),  (8,  0),  (0,  -  6). 

Ans.   x2  +  ^2  _  3  a;  _^  (5  y  _  0^ 

(c)  passes  through  the  points  (4,  0),  (—  2,  5),  (0,  —  3). 

Ans.    19x2 +-19^2^  2x- 47  2/ -  812  =  0. 

(d)  passes  through  the  points  (3,  5)  and  (—3,  7)  and  has  its  center  on 
the  JT-axis.  Ans.    x2  +  y2  _|_  4  j.  _  46  =  Q. 

(e)  passes  through  the  points  (4,  2)  and  (—6,  —  2)  and  has  its  center  on  "* 
the  F-axis.  Ans.   x2  +  2/2  +  5  2/  -  30  =  0. 

(f)  passes  through  the  points  (5,  —  3)  and  (0,  6)  and  has  its  center  on 
the  line2x -3?/ -  6  =  0.  Ans.    3x2  +  82/2  -  114x  -  642/ +  276  =  0. 

(g)  has  the  center  ( —  1,  —  5)  and  is  tangent  to  the  JT-axis. 

Ans.  x2  +  ^2  _^  2  X  +  10  2/  +  1  =  0. 
(h)  passes  through  (1,  0)  and  (5,  0)  and  is  tangent  to  the  F-axis. 

^715.    x2  +  2/^-6x±2V5y  +  5  =  0. 
(i)  passes  through  (0,  1),  (5,  1),  (2,  -  3). 

Ans.    2x2  +  2^2_iox  +  2/-3  =  0. 
(j)  has  the  line  joining  (3,  2)  and  (—  7,  4)  as  a  diameter. 

Ans.    x2  +  2/^  +  4  X  -  6  2/  -  13  =  0. 
(k)  has  the  line  joining  (3,  —  4)  and  (2,  —  5)  as  a  diameter. 

Ans.    x2  +  2/2  -  5  X  +  9  y  +  26  =  0. 
(1)  which  circumscribes  the  triangle  formed  by  x  —  6  =  0,  x4-22/  =  0, 
and  X  -  2  2/  =  8.  Ans.  "2  x2  +  2 y2  _  21  x  +  8  ?/  +  60  =  0. 

(m)  passes  through  the  points  (1,  —  2),  (—  2,  4),  (3,  —  6).     Interpret  the 
result  by  the  Corollary,  p.  98. 

(n)  is  inscribed  in  the  triangle  formed  by4x  +  32/  —  12  =  0,  2/  —  2  =  0, 

tx  -  10  =  0.  Ans.    36  x2  +  36  2/2  -  516  x  +  60  2/  +  1585  =  0. 


4.  Plot  the  locus  of  x2  +  2/2  -  2  x  +  4  7/  +  A:  =  0  for  A;  =  0,  2,  4,  5  -  2,  -  4, 
8.     What  values  of  k  must  be  excluded ?  Ans.   lc>h. 


136  ANALYTIC  GEOMETRY 

5.  What  is  the  locus  of  x"^  +  y^  +  Dx  +  Ey  -{-  F  =  0  ii  DandE  are  fixed 
and  F  varies  ? 

6.  For  what  values  of  k  does  the  equation  A"^  +  y^  —  4x  +  2 ky  +  10  =  0 
have  a  locus  ?  Ans.   k>  -{■  V6  and  A;  <  —  V6. 

7.  For  what  values  of  k  does  the  equation  x^  +  y^  -\-  kx  +  F  =  0  have  a 
locus  when  (a)  F  is  positive ;  (b)  F  is  zero  ;  (c)  F  is  negative  ? 

Ans.    (a)  ^ > 2 Vf and  k<-2 ^F ;  (b)  and  (c)  all  values  of  k. 

8.  Find  the  number  of  point-circles  represented  by  the  equation  in 
problem  7.  Ans.    (a)  two ;  (b)  one  ;  (c)  none. 

9.  Find  the  equation  of  the  circle  in  oblique  coordinates  if  w  is  the  angle 
between  the  axes  of  coordinates. 

Ans.    {x  -  a:)2  +  (?/  -  /3)2  +  2  (x  —  a)  {y  -  ^)  cos  w  =  r'^. 

10.  Write  an  equation  representing  all  circles  with  the  radius  6  whose 
centers  lie  on  the  X-axis ;  on  the  Y-axis. 

11.  Find  the  number  of  values  of  k  for  which  the  locus  ot^ 

(a)  x^  +  y"^  -}-  4:kx  -  2 y  +  6 k  =  0, 

(b)  x^  +  y^  +  4:kx  -  2y  -  k  =  0, 

(c)  x2  -f  2/2  +  4  fcx  -  2  2/  +  4  A:  =  0 

is  a  point-circle.  Ans.    (a)  two ;  (b)  none ;  (c)  one. 

12.  Plot  the  circles  a;^  +  ?/2  +  4 x  -  9  =  0,  x'^  +  y^  -  ix  -  9  =  0,  and 
x2  -f-  2/2  _|_  4x  _  9  +  A:(x2  +  2/2  _  4x  -  9)  =  0  for  A:  =  ±  1,    ±3,    ±  i,    -  5, 

—  ^.     Must  any  values  of  k  be  excluded  ? 

13.  Plot  the  circles  x2  +  2/2  +  4x  =  0,  x2  +  2/2  _  4^;  =  0,  and  x2  +  y2  +  4x 
-f  A;  (x2  -|-  2/2  —  4  x)  =  0  for  the  values  of  k  in  problem  12.  Must  any  values 
of  A:  be  excluded  ? 

14.  Plot  the  circles  x2 -1- ?/2  +  4x  +  9  =  0,  x2  +  ?/2  -  4x  +  9  =  0,  and 
x2  +  2/2  +  4x4-9  + A:(x2  + 2/2-4x4- 9)  =  0   for   A;  =  -  3,    -  |,    -5,    -  |, 

—  I,  —  f ,   —  1.     What  values  of  k  must  be  excluded  ? 

'  58.  Systems  of  circles.    An  equation  of  the  form 

x^-\-y'^  +  Dx  +  Ey  +  F=0 
will  define  a  system  of  circles  if  one  or  more  of  the  coefficients 
contain  an  arbitrary  constant.     Thus  the  equation 

x^  -\-  7/^  —  r^  =  0 
represents  the  system  of  concentric  circles  whose  centers  are  at 
the  origin.     Very  interesting  systems  of  circles,  and  the  only 
systems  we  shall  consider,  are  represented  by  equations  analogous 
to  (XIII),  p.  119. 


THE  CIRCLE  137 

Theorem  III.    Given  two  circles,  ' 

Ci :  ^2  +  7/2  +  D^x  +  E^y  +  i^'i  =  0 
and  C^:x^  -{-if  -{-  D^x  +  E^y  +  F^  =  0; 

then  the  locus  of  the  equation 

(III)  oc^  +  y^  +  n^oc  +  Eiu  +  F^ 

+  l^iic''  +  2/^  +  i>2^  +  E^y  +  JPa)  =  O 

t5  (Z  circle  except  when  k  =  —  X.     In  this  case  the  locus  is  a  straight 
line. 

Proof.  Clearing  the  parenthesis  in  (III)  and  collecting  like 
terms  in  x  and  y,  we  obtain 

{l  +  l<)x^-\-{l-\-k)y^  +  {Dy^kR,)x  +  {E,  +  kE^)y  +  {F,  +  kF,)  =  0. 

Dividing  by  1  +  ^  we  have 

"^  +y  ^  ij^k  ^^  1  +  /^  ^     i^rk     ^• 

The  locus  of  this  equation  is  a  circle  (Theorem  I,  p.  131).  If, 
however,  k  =  —  1,  we  cannot  divide  by  1  +  A^.  But  in  this  case 
equation  (III)  becomes 

(A  -  D,)x  +  (E,  -E,)y+  (F,  -  F,)  =  0, 

which  is  of  the  first  degree  in  x  and  y.     Its  locus  is  then  a 
straight  line  called  the  radical  axis  of  Ci  and  C^.  q.e.d. 

Corollary  I.  The  center  of  the  circle  (III)  lies  upon  the  line 
joiiiing  the  centers  of  Ci  and  Cg  and  divides  that  line  into  seg- 
ments whose  ratio  is  equal  to  k. 

For  by  Theorem  I  (p.  131)  the  center  of  Ci  is  Pi(  -  — i.   -  -^  )  and  of  C-i  is 
f      D2         Ei\  \       2  2  / 

P2  ( » j  •    The  point  dividins?  P1P2  into  segments  whose  ratio  equals  k 

is  (Theorem  VII,  p.  39)  the  point      — »  -—— >  or, 

L—  1-f-A^  l-rfl!  —J 

.      ,.,  .       (      Dx  +  kDi         Ei-{-kE2\       ^  .  .    .    ^^  ^        .  .tttn 

sjmphfymg,  (^  -  ^  >   -  -^  j '  which  is  the  center  of  (III). 


138 


ANALYTIC  GEOMETRY 


Corollary  II.    The  equation  of  the  radical  axis  of  C-^  and  C^  is 

(A  -D,)x+  (E,  -  E,)y  +  (Fi  -  F,)  =  0. 

Corollary  III.    The  radical  axis  of  two  circles  is  perpendicular  to 
the  line  joining  their  centers. 

Hint.   Find  the  line  joining  the  centers  of  C^  and  C2  (Theorem  VII,  p.  97)  and  show 
that  it  is  perpendicular  to  the  radical  axis  by  Corollary  III,  p,  87. 

.  The  system  (III)  may  have  three  distinct  forms,  as  illustrated 
in  the  following  examples.  These  three  forms  correspond  to  the 
relative  positions  of  C^  and  Cg,  which  may  intersect  in  two  points, 
be  tangent  to  each  other,  or  not  meet  at  all. 

Ex.  1.    Plot  the  system  of  circles  represented  by 

x2  +  2/2  4-  8x  -  9  +  ^-(x2  +  ?/2  -  4a;  -  9)  =  p^,   • 


Solution.    The  figure  shows  the  circles 

a:2  +  7/2  +  8 X  -  9  =  0  and  x'^  -h  y^  -  4:X  -  9  =  0 

plotted  in  heavy  lines  and  the  circles  corresponding  to 

A;  =  2,  5,  1,  4,-4,  -  f ,  and  -  I ; 

these  circles  all  pass  through  the  intersection  of  the  first  two. 

The  radical  axis  of  the  two  circles  plotted  in  heavy  lines,  which  corre- 
sponds to  k  =  —  1,  is  the  F-axis. 


THE  CIRCLE 


139 


Ex.  2.    Plot  the  system  of  circles  represented  by 

x2  +  ?/2  -f  8x  +  k{x^  +  2/2  -  ix)  =  0. 


- 

^ 

-= 

^ 

Y, 

^ 

-- 

-- 

N 

-- 

/ 

/ 

y 

\ 

/^rVJ 

^ 

s 

\ 

/ 

/ 

/ 

/' 

^\ 

r 

N 

\, 

^ 

1 

( 

/ 

/ 

^\ 

'P^ 

Ni 

\ 

\ 

\l 

( 

r  A 

^3 

\ 

X 

\ 

^■^ 

fc3 

J 

X 

\ 

\ 

\ 

\ 

^!^ 

^Ki 

/' 

1 

/I 

\ 

\ 

\^ 

^f 

\^ 

^-^ 

y 

/ 

/ 

\ 

\ 

\ 

^ 

\- 

■^ 

-^ 

y 

/ 

- 

^ 

-^ 

/ 

r 

\. 

^> 

^ 

^ 

-- 

Solution.    The  figure  shows  the  circles 

x^  +  y^  +  Sx  =  0  and  x^  +  y'^-4:X  =  0 
plotted  in  heavy  lines  and  the  circles  corresponding  to 

A:  =  2,  3,  I,  5,  1,  1,  -  7,  i,  -  4,  -  3,  and  -  f 

These  circles  are  all  tangent  to  the  given  circles  at  their  point  of  tangency. 
The  locus  for  Z:  =  2  is  the  origin. 

Ex.  3.    Plot  the  system  of  circles  represented  by 

'       x2  +  2/2_iOx  +  9  +  A;(x2  +  2/2  +  8cc  +  9)  =  0. 


Solution.    The  figure  shows  the  circles 

x2  +  y2  _  10 X  +  9  =  0  and  x2  +  ?/2  _^.  gx  +  9 
plotted  in  heavy  lines  and  the  circles  corresponding  to 
Jc  =  h  17,  h  -  10,  -  tV.  and  -  V- 


140  ANALYTIC  GEOMETRY 

These  circles  all  cut  the  dotted  circle  at  right  angles,  as  will  be  shown 
later.  For  k  =  f  the  locus  is  the  point-circle  (3,  0),  and  for  A:  =  8  it  is  the 
point-circle  (—3,  0). 

In  all  three  examples  the  radical  axis,  for  which  A;  =  —  1,  is  the  F-axis. 

Theorem  IV.     When  the  circles 

Ci  :  ^2  _^  2/'  +  A^  -h  E,ij  +  Fi  =  0 
and  C^  :  x^  -\-  y^  -\-  D^x  -f  Ec^y  +  F^  =  0 

intersect  in  two  points  P^  (xi,  y^)  and  P^  (x^,  y^,  then  the  system 
of  circles  represented  by 
a;2  +  y2  _,_  ^^^  _j_  ^^y  _j_  2^^  _p  ;^(^2  _^  y2  _^  ^^^  ^  ^^^  +  ^2)  =  0 

consists  of  all  circles  passing  through  P-^  and  P^. 

Proof.  First,  every  circle  of  the  system  passes  through  Pj  and 
Pa-     I'or,  since  Pj  lies  on  C^  and  Cg,  we  have 

^1  +  2/1'  +  A^i  +  A2/1  +  i^i  =  0 
and  x^  -^  2/i2  ^  ^^^^  ^  ^^^^  +  P2  =  0. 

Multiply  the  second  equation  by  h  and  add  to  the  first;  this 
gives 

which  is  the  condition  that  P^  lies  on  any  circle  of  the  system. 
In  the  same  manner  we  can  show  that  every  circle  of  the  system 
passes  through  P^. 

In  the  second  place,  every  circle  which  passes  through  P^  and 
P2  is  in  the  system.     For  any  such  circle  is  determined  by  Pi,  Pg 
and  a  point  P3  (ccg,  y^)  not  on  the  line  P^P^-     Then  if  P^  lies  on  a 
circle,  of  the  system,  we  have 
^3'  +  2/3'  +  D^x^  +  E^y^  +  i^i  +  /v  (rrg^  +  y,^  +  D^x^  +  i^J^T/g  +  P2)  =  0, 

and  hence  ^  ^  _  ^a^  +  ^a^  +  A^3  +  A.Vs  +  ^1. 
a^s'  +  2/3'  +  Aa^s  +  -E^22/3  -f  P2 
That  is,  a  value  of  k  can  be  determined  so  that  the  corresponding 
circle  passes  through  P3.  Since  P3  is  any  point  not  on  P^P^, 
that  circle  is  any  circle  which  passes  through  Pi  and  P^;  and 
hence  every  circle  which  passes  through  Pj  and  Pg  belongs  to 
the  system.  q.e.d. 


THE  CIRCLE  141 

Corollary.  The  radical  axis  of  two  iiitersecting  circles  is  their 
common  chord. 

In  like  manner  we  may  prove 
Theorem  V.    When  the  circles 

Ci :  ^2  +  2/2  +  D^x  +  J5:iy  +  Fi  =  0 
and  Ca  :  cc^  +  2/^  +  D^^  +  E^y  -^  F^  =  0 

are  tangent  at  the  j^oint  Py  (xi,  t/j),  then  the  system  of  circles  repre- 
sented hy 

^'  4-  2/'  +  Aa^  +  E,y  J^F^-\-k{x^  +  t/  +  D^x  +  E^y  +  ^2)=  0 
consists  of  all  circles  tangent  to  C^  and  C^  at  Pj. 

These  theorems  show  how  to  construct  the  circles  of  the  system 
in  case  Ci  and  Cg  intersect  or  are  tangent,  but  there  is  no  analo- 
gous theorem  if  Ci  and  Cg  do  not  intersect.  In  what  follows  we 
shall  consider  a  method  which  applies  to  all  three  cases. 

Theorem  VI.  The  equation  of  the  sy stein  (III),  (p.  137),  may  he 
tvritten  iii  the  form 

(VI)  ic2   +  y2  _,_  j^f^  _,_  2?^  ^  Q^ 

where  k'  is  an  arbitrary  constant,  if  the  axes  of  x  and  y  be  respec- 
tively chosen  as  the  line  of  centers  and  the  radical  axis  of  Ci  and  Cg. 

Proof  No  matter  how  the  axes  be  chosen,  the  equations  of 
Ci  and  C2  have  the  forms 

C^:x^  +  7/  +  D,x  +  Fiy  +  Fi  =  0 
and  C2  :  £c2  +  ?/2  4-  D^x  +  E^y  +  F^  =  0. 

If  the  centers  of  Ci  and  C\  lie  on  the  Z-axis,  then  Fi  =  0  and 
F2  =  0  (Corollary,  p.  131).  The  equation  of  the  radical  axis 
(Corollary  II,  p.  138)  then  becomes 

{D,-D,)x^{F,-F,)=0. 

If  this  line  is  the  F-axis,  whose  equation  is  x  =  Q,  we  must 
have  Fi  —  F2  =  0,  and  "hence  F^  =  F^.  Substituting  F  for  Fj 
and  Fa  and  setting  E^  =  0  and  F2  =  0  in  (III),  we  obtain 

x"^  -{-  y""  +  D^x  -{-  F  -{-  k  (.t2  +  t/^  +  D^x  +  F)  =  0. 


142  ANALYTIC  GEOMETRY 

Collecting  like  powers  of  x  and  y  and  dividing  by  1  +  yfc,  we 
obtain 

x^  ^-tf  ^ ic  +  F  =  0. 

X  -]-  ki 

The  coefficient  of  a;  changes  with  k  and  may  be  denoted  by  a 
single  letter;  if  we  set 

A  +  hP, 
1-^k     -"' 
we  obtain  equation  (VI).  q.e.d. 

Corollary.  The  centers  of  the  circles  of  the  system  (YI)  lie  on 
the  X-axis. 

The  study  of  the  system  of  circles  (III),  p.  137,  may  then  be 
effected  by  the  study  of  the  system  (VI),  whose  equation  is  in  a 
simpler  form  than  that  of  (III). 

Theorem  VII.    If  r'  is  the  radius  of  that  circle  of  the  system 

X^  +  2/2  +  k'x  +  F  =  0 

whose  center  is  (a\  0),  then 


k'^  —  4:  F 
Proof    For  by   Theorem  I   (p.   131)  we  have  r'^  = 

and  a'2  =  -— •     Hence  r'^  =  a'"^  —  F. 
4 

Corollary  I.  When  F  is  negative,  r'  is  the  hypotenuse  of  a  right 
triangle  ivhose  legs  are  a!  and  V—  F .^ 

Corollary  II.    When  F  is  zero,  then  r'  =  a'. 

Corollary  III.  When  F  is  positive,  a'  is  the  hypotenuse  of  a  right 
triangle  ivhose  legs  are  r'  and  '\F . 

We  may  readily  construct  circles  of  the  system  (VI)  by  the 
use  of  these  corollaries.  With  the  preliminary  remark  that  the 
centers  of  all  of  the  circles  of  the  system  lie  on  the  Z-axis  (by 
the  Corollary),  we  shall  consider  the  three  cases  separately. 

*  When  F  is  negative,  —F  is  positive,  and  hence  y/-F  is  a  real  number. 


THE  CIRCLE 


143 


Case  I.  F  <  0.  In  this  case  r'^  =  a'^  —  F  is  positive  for  all 
real  values  of  a',  and  hence  every  point  on  the  A'-axis  may  be 
used  as  the  center  of  a  circle  belonging  to  the 
system. 

On  OF  lay  off  OA  =  V— i^.  With  any 
point  P'  on  the  Z-axis  as  center  and  with. 
P'A  as  a  radius,  describe  a  circle ;  this  circle 
will  belong  to  the  system.  For  let  OP'  =  a'; 
then  P'A  =  r'  by  Corollary  I.  The  system 
is  then  composed  of  all  circles  whose  centers 
lie  on  the  Z-axis  which  pass  through  A  (0,  +  V—  F).  It  is  evi- 
dent that  the  circles  will  also  pass  through  B  (0,  —  V—  F). 
Case  II.  F  =0.  In  this  case  r''^  =  a'^,  and  hence  all  points 
on  the  Z-axis  may  be  used  as  centers.  Further, 
the  circles  of  the  system  will  all  pass  through 
the  origin  (Theorem  YI,  p.  73).  Hence  the 
circle  whose  center  is  any  point  P'  on  the 
Z-axis  and  whose  radius  is  P'O  will  belong  to 
the  system.  It  is  evident  that  all  of  the  cir- 
cles of  the  system  are  tangent  to  the  F-axis  at  the  origin  and 
also  to  each  other. 

Case  III.  F>  0,  In  this  case  r'^  =  a'^  —  F  is  positive  only 
when  a'  is  numerically  greater  than  \F,  and  hence  points  on  the 
Z-axis  for  which  a'  is  numerically  less 
than  V^  cannot  be  used  as  centers. 
With  0  as  a  center  and  with  Vf  as  a 
radius,  describe  a  circle,  the  dotted  circle 
in  the  figure.  Let  P'  be  any  point  on  the 
Z-axis  outside  of  this  circle.  Draw  P'A 
tangent  to  the  dotted  circle.  With  P'  as 
center  and  P'A  as  radius,  describe  a  circle;  this  circle  will  belong 
to  the  system.  For  let  P'0  =  a'',  then,  since  OA  =  Vf,  and  since 
^  is  a  right  angle,  P'A  =r'  by  Corollary  III.  Two  intersecting 
jircles  whose  tangents  at  a  point  of  intersection  are  perpendicu- 
lar are  said  to  be  orthogonal ;  hence  the  system  is  composed  of  all 
circles  whose  centers  are  on  the  Z-axis  which  cut  the  dotted  circle 


144  ANALYTIC  GEOMETRY 

orthogonally.  If  P'  falls  at  C  or  D,  the  radms  will  be  zero ; 
that  is,  the  point-circles  C  and  D  belong  to  the  system  and  are 
called  its  limiting  points.     Hence 

Theorem  VIII.    The  circles  of  the  system  represented  by 

a;2  4. 2/2  +  h'x  -\-F=0 

have  their  centers  on  the  X-axis,  and 


(a)  pass  through  (0,  +  V—  F)  and  (0,  —  V—  F)  if  F  is 
negative  ; 

(b)  are  tangent  to  each  other  at  the  origin  ifF=  0; 

(c)  are  orthogonal  to  the  circle  x^  +  y'^  =  F  if  F  is  positive. 

The  constructions  given  in  the  proof  were  used  in  drawing  the 
figures  on  pages  138  and  139. 

It  is  evident  from  the  figures,  and  can  be  proved  analytically, 
that  there  are  no  point-circles  if  F  is  negative,  that  there  is  one 
point-circle  if  F  is  zero,  and  that  there  are  two  if  F  is  positive. 

59.  The  length  of  the  tangent. 

Theorem  IX.    Given  a  point  Pi  (xi,  y-^  and  the  circle 

C  -.x"^ +  !/-{- Dx-\-Ey  -^F^O, 

then  the  product  of  any  secant  through  P^  and  its  external  s.eg- 
Tnent  is 

Proof    Let  the  equations  of  any  line  through  Px  be  (Theorem 

XV,  p.  124) 

x  =  Xx  +  p  cos  a, 
2/  =  2/1  +  P  cos  /?. 

Then  if  the  point  (x,  y)  or  (x^  -f 
/>j    p  cos  a,  yi  +  p  cos  (3)   lies   on  C,  we 
have  (Corollary,  p.  53) 

(xi  +  p  cos  a)2  +  (?/i  -f  p  cos  (3y 

+  D(xi-\-  p  cos  a)  4-  -^  (2/1  +  f>  cos  l3)-\-F=0. 


THE  CHICLE  146 

Simplifying,  arranging  according  to  powers  of  p,  and  using 
(1),  p.  123,  we  have 

p2  +  p  [(2  Xi  +  D)  cos  «  +  (2  2/1  +  £)  cos  y8] 

+  ^i'  +  2/i'  +  Dx^  +  Eij,  +  Fi  =  0. 

The  roots  of  this  quadratic  are  the  lengths  of  the  secant  PA 
and  its  external  segment  P1P2'  Hence  the  product  of  I\Ps  and 
P1P2  is  (Theorem  I,  p.  3) 

As  this  expression  does  not  contain  cos  a  or  cos  /3  it  is  imma- 
terial in  what  direction  the  secant  be  drawn.  q.e.d. 

Corollary.    The  square  of  the  length  of  the  tangent  from  P^  to  C 

is  given  by  (IX). 

For  when  the  secant  swings  around  on  Pi  until  it  becomes  tangent  to  C,  P1P3 
and  P1P2  both  become  equal  to  P1P4. 

Theorem  X.  The  ratio  of  the  squares  of  the  lengths  of  the  tan- 
gents di'awn  from  any  point  of  the  circle 

Cj,:x'-\-y''^-  D,x  +  E,y  +  ^i 

+  k(x^  +  2/'  +  D2X  +  E,y  +  P2)  =  0 
to  the  circles 

C,:x'  +  y''-\-  D,x  -^E,y-]-F,  =  0 

and  C^-.x^  +  2/  -\-  D^x  +  E^y  4-  Pg  =  0 

is  constant  and  is  equal  to  —  k. 

Proof    Let  Pi(a:i,  y^  be  any  point  on  Cj^.     Then 
^i^  +  Vi  +  A^i  +  ^i2/i  +  ^1  +  ^  (^1'  +  Vi  +  D^^i  +  E^y,  +  P2)  =  0. 

Dividing  by  the  parenthesis  and  transposing,  we  obtain 
^i"  +  yi'  +  A^i  +  -gi  3/1  +  Pi  ^      j^ 

^l'  +  2/l'  +  A^l  +  P2  2/l  +  ^2 

By  the  Corollary  the  numerator  of  this  fraction  is  the  square 
of  the  length  of  the  tangent  from  Pj  to  C^,  and  the  denominator 
is  the  square  of  the  length  of  the  tangent  from  Pj  to  Cg.  Hence 
the  ratio  of  the  squares  of  the  lengths  of  those  tangents  is  con- 
stant and  equal  to  —  A;.  q.e.d. 


146  Ai^ALYTIC  GEOMETRY 

Corollary  I.   The  locus  of  a  point  from  which  the  ratio  of  the 

squares  of  the  lengths  of  the  tangents  to  the  circles  C^  and  C^  is 

constant  and  equal  to  —  k  is  the  circle  C^. 

Theorem  X  proves  only  one  part  of  the  Corollary.  It  remains  to  be  proved 
that  all  points  such  that  the  ratio  of  the  squares  of  the  lengths  of  the  tangents  from 
these  points  to  C\  and  (J%  equals  —  Ic  lie  on  C'^.. 

Corollary  II.  The  locus  of  points  from  which  tangents  to  two 
circles  are  equal  is  the  radical  axis  of  those  circles. 


PROBLEMS 

1.  By  means  of  Theorem  VIII  plot  the  following  systems  of  circles. 

(a)  ic2  +  2/2  +  4a;  -  1  +  A;(x2  +  2/2  -  2x  -  1)  =  0. 

(b)  x2  +  2/2  -f  4x  +  1  +  A;(x2  +  2/2  -  2x  +  1)  =  0. 

(c)  x2  +  2/2  +  4x  +  A;(x2  +  y^-2x)  =  0. 

(d)  x2  +  2/2  +  2x  -  4  +  fc (x2  +  2/2  +  Cx  -  4)  =  0.  IP 

(e)  x2  +  2/2  +  2x  +  9  +  fc (x2  +  2/2  -  4x  +  9)  =  0. 

(f )  x2  +  2/2  -  6  X  +  A;  (x2  +  2/2  +  8  x)  =  0. 

2.  Find  the  lengths  of  the  tangents  from  the  point 

(a)  (5,  2)  to  the  circle  x2  +  ^2  _  4  =  q. 

(b)  (-  1,  2)  to  the  circle  x^  +  y^-6x-2y  =  0. 

(c)  (2,  5)  to  the  circle  2x2  +  2  2/2  +  2x  +  42/-l=0. 

(d)  (1,  2)  to  the  circle  x^-\-y^  =  25. 

What  does  the  imaginary  answer  in  (d) mean  ?  Ans.  Point  is  within  the  circle. 

3.  Determine  the  nature  of  the  following  systems. 

(a)  x2  +  2/2  +  2  X  -  4  2/  +  A;  (x2  +  y2  _  2  X  +  4  y)  =  0. 

(b)  x2  +  2/2  +  4  X  -  y  +  fc  (x2  +  2/2  -  4  X  +  2/  -  4)  =  0. 

(c)  x2  +  2/^+  2 X  -  4 2/  +  1  +  A; (x2  +  2/2  -  2 X  +  4  y  +  1)  =  0. 

4.  Find  the  equation  of  the  circle  passing  through  the  intersections  of  the 
circles  x^  +  y^  —  1  =  0  and  x^  -{-  y^  +  2x  =  0  which  passes  through  the  point 
(3,2).  ^ns.   7x2 +  7  2/2- 24x- 19  =  0. 

6.  Find  the  equation  of  the  circle  passing  through  the  intersections  of 
x2  -I-  2/2  —  6  X  =  0  and  x2  +  2/2  —  4  =  0  which  passes  through  (2,  —  5). 

Ans.    x2 +  2/^-3x-2  =  0. 

6.  Find  the  equation  of  that  circle  of  the  system  x^ -\- y^  —  Ax  —  3 
+  A; (x2  +  2/'"^  —  4 2/  —  3)  =  0  whose  center  lies  on  the  line  x  —  y  —  4:ih=X). 

Ans.   X2  +  2/2 -6x  +  2|/-3='0- 


THE  CIRCLE  147 

7.  Find  the  equation  of  the  circle  passing  through  the  intersections  of 
x^  -{■  y^  —  i X  +  2y  =  0  and  x^  +  ^z^  —  2y  —  4  =  0  whose  center  lies  on  the 
line  2x  +  4y-l  =  0.  Ans.   x'^  +  y^  -  3x  +  y  -  1  =  0. 

8.  Find  the  equations  of  the  circles  passing  through  the  intersections  of 
aj2  _|_  y2  _  4  —  0  and  x^  +  y^  +  2x  —  S  =  0  whose  radii  equal  4. 

Ans.   x^-\-y^  —  6x-7  =  0  and  x^ -{- y^  -\- 8x  =  0. 

9 .  Find  the  radical  axes  of  the  circles  x^  +  y2 — 4  x = 0,  x^ + ?/2 + 6  x  —  8  ?/ = 0, 
and  x2  +  2/2  +  6x  —  8  =  0  taken  by  pairs,  and  show  that  they  meet  in  a  point. 

10.  Find  the  radical  axes  of  the  circles  x^-]-y^-9=0,  3x2+3 2/2-6 x+82/ 
— 1=0,  and  x2+2/2_)_8  y=0  taken  by  pairs,  and  show  that  they  meet  in  a  point. 

11.  Show  that  the  radical  axes  of  any  three  circles  taken  by  pairs  meet 
in  a  point. 

12.  By  means  of  problem  11  show  that  a  circle  may  be  drawn  cutting  any 
three  circles  at  right  angles. 

13.  By  means  of  problem  11  prove  that  if  several  circles  pass  through  two 
fixed  points  their  chords  of  intersection  with  a  fixed  circle  will  pass  through 
a  fixed  point. 

14.  The  square  of  the  tangent  from  any  point  Pi  of  one  circle  to  another 
is  proportional  to  the  distance  from  the  radical  axis  of  the  two  circles  to  Pi. 

15.  If  Ci  and  C^  (Theorem  III)  are  concentric,  then  all  the  circles  of  the 
systeni  (III)  are  concentric. 

16.  Show  that  when  Ci  and  C2  (Theorem  III)  are  concentric  the  equation 
of  the  system  (III)  cannot  be  written  in  the  form  given  in  Theorem  VI. 

17.  Show  that  the  radical  axis  of  any  pair  of  circles  in  the  system  (III) 
is  the  same  as  the  radical  axis  of  Ci  and  C2. 

18.  How  may  problem  11  be  stated  if  the  three  circles  are  point-circles  ? 

MISCELLANEOUS   PROBLEMS 

1 .  Find  the  equation  of  the  circle  which  circumscribes  the  triangle  formed 
byx  +  2?/  =  0,  3x-2y  =  6,  and  x  -  y  =  6. 

2.  Find  the  equation  of  the  circle  inscribed  in  the  triangle  in  problem  1. 

3.  Find  the  angle  between  the  radii  of  the  circles  x^  -\-y^  =  26  and 
x^  -\-y^  —  \6x-hS9  =  0  which  are  drawn  to  a  point  of  intersection. 

Hint.    Find  the  radii,  the  length  of  the  line  of  centers,  and  apply  17,  p.  20. 

4.  Find  the  angle  between  the  radii  of  the  circles  x^  +  y^  +  DiX  +  Eiy 
+  Pi  =  0  and  x2  +  2/2  ^  j)^x  +  P22/  +  P2  =  0  which  are  drawn  to  a  point  of 
intersection. 


148  ANALYTIC  GEOMETRY 

5.  Find  the  condition  that  the  angle  in  problem  4  should  be  a  right  angle. 

6.  Show  that  an  angle  inscribed  in  a  semicircle  is  a  right  angle. 

7.  Prove  that  the  perpendicular  dropped  from  a  point  on  a  circle  to  a 
diameter  is  a  mean  proportional  between  the  segments  of  the  diameter. 

8.  If  w  is  the  angle  between  the  oblique  axes  OX  and  OY,  then  the 
locus  of  x2  +  2  cos  (oxy  +  ?/2  +  Dx  +  jE"?/  +  F  =  0  is  a  circle. 

9.  Given  a  circle  C:x2+2/2+Zte+^2/+ 2^=0  and  a  line  i:J.x+%+C=0; 
show  that  the  system  of  curves  x"^ -\-  y^  -\-  Bx  +  Ey  +  F  +  k  {Ax  +  By  -j-C)=0 
consists  of  all  circles  whose  centers  lie  on  the  line  through  the  cent.er  of  C 
perpendicular  to  L. 

10.  Find  the  radical  axis  of  any  two  circles  of  the  system  in  problem  9. 

1 1 .  Find  a  geometric  interpretation  of  k  in  the  equation  in  problem  9. 

12.  What  does  the  equation  of  the  system  in  problem  9  become  if 

(a)  the  Y-axis  is  the  line  L  and  the  X-axis  passes  through  the  center  of  C  ? 

(b)  the,  origin  is  the  center  of  C  and  the  Y-axis  is  chosen  parallel  to  L? 

13.  Show  how  to  construct  the  circles  of  the  system  x^+y^—r^-{-k{x—a)=0 
when  (a)  r < a ;  {h)  r  =  a;  and  (c)  r>a.  -i 

14.  Show  that  the  discriminant  of  (III)  is 

r2^k^  -  {(P'  -  r^  -  r2^)  ^  +  r^ 
(1  +  A:)2  ' 

where  ri  is  the  radius  of  Ci,  r^  of  C2,  and  d  is  the  length  of  the  line  joining 
the  centers  of  C\  and  C2. 

15.  From  problem  14  show  that  if  there  are  no  point-circles  in  (III),  then 
Ci  and  C2  intersect ;  if  there  is  one  point-circle  in  (III),  then  C\  and  C2  are 
tangent ;  if  there  are  two  point-circles  in  (III),  C\  and  Ci  do  not  intersect. 


CHAPTER  VI 


POLAR   COORDINATES 


60.  Polar  coordinates.  In  this  chapter  we  shall  consider  a 
second  method  of  determining  points  of  the  plane  by  pairs  of 
real  numbers.  We  suppose  given  a  fixed  point  0,  called  the 
pole,  and  a  fixed  line  OA,  passing  through 
0,  called  the  polar  axis.  Then  any  point 
P  determines  a  length  OP  =  p  and  an 
angle  A  OP  =  0.  The  numbers  p  and  0 
are  called  the  polar  coordinates  of  P.  p  is 
called  the  radius  vector  and  6  the  vectorial  \ 

angle.     The  vectorial  angle  0  is  positive  \ 

or  negative  as  in  Trigonometry  (p.  18).  "^ 

The  radius  vector  is  positive  if  P  lies  on  the  terminal  line  of  6, 
and  negative  if  P  lies  on  that  line  produced  through  the  pole  0. 
Thus  in  the  figure  the  radius  vector  of  P  is  positive,  and  that  of  P'  is  negative. 

It  is  evident  that  every 
pair  of  real  numbers  (p,  0) 
determines  a  single  point, 
which  may  be  plotted  by 
the 

Rule  for  plotting  a  point 
whose  polar  coordinates 
(p,  0)  are  given. 

First  step.  Construct  the 
terminal  line  of  the  vecto- 
rial angle  0,  as  in  Trigo- 
nometry. 

Second  step.  If  the  radius 
vector  is  positive,  lay  off  a 
length  OP  =  p  on  the  terminal  line  of  6;  if  negative,  produce  the 

149 


150  ANALYTIC  GEOMETRY 

terminal  line  through  the  pole  and  lay  off  OP  equal  to  the  numer- 
ical value  of  p.     Then  P  is  the  required  point. 

In  the  figure  on  p.  149  are  plotted  the  points  whose  polar  coordinates  are 

(''-f>(^-T>(-'X><«-).-(^.-f> 

Every  point  P  determines  an  infinite  number  of  pairs  of  numbers  (p,  6). 

The  values  of  d  will  differ  by  some  mul- 
tiple of  7t,  so  that  if  0  is  one  value  of  6  the 
others  will  be  of  the  form  0  +  kTt,  where  k  is 
a  positive  or  negative  integer.  The  values 
of  p  will  be  the  same  numerically,  but  will 
be  positive  or  negative,  if  P  lies  on  OB, 
according  as  the  value  of  6  is  chosen  so  that 
OB  or  OC  is  the  terminal  line.  Thus,  if 
OB  =  p  the  coordinates  of  B  may  be  written 
in  any  one  of  the  forms  (p,  0),  (—  p,  tt  +  0), 
(/),  2 TT  +  0),  {—  p,  (p—7t),  etc. 

Unless  the  contrary  is  stated,  we  shall  always  suppose  that  6  is 
positive,  or  zero,  and  less  than  2  ir ;  that  is,  0  <  ^  <  2  tt. 

PROBLEMS 

1.  Plot  ..ep„lnu(4.|),(e  -),(-.,  VO.(^,f).(-/-^). 

(5,  7t). 

2.  Plot  the  points  (6,  ±j),  (-2,  ±  |),  (3,  7t),  (-4,  tt),  (6,  0), 
(-6,  0).  \  4/     \  2/ 

3.  Show  that  the  points  (/>,  6)  and  (p,  —  d)  are  symmetrical  with  respect 
to  the  polar  axis. 

4.  Show  that  the  points  (p,  6),  (—  p,  0)  are  symmetrical  with  respect  to 
the  pole. 

5.  Show  that  the  points  {—  p,  7t  —  6)  and  (p,  6)  are  symmetrical  with  respect 
to  the  polar  axis. 

61.  Locus  of  an  equation.  If  we  are  given  an  equation  in  the 
variables  p  and  6,  then  the  locus  of  the  equation  (p.  59)  is  a  curve 
such  that : 

1.  Every  point  whose  coordinates  (p,  6)  satisfy  the  equation  lies 
on  the  curve. 

2.  The  coordinates  of  every  point  on  the  curve  satisfy  the 
equation. 


POLAR  COORDINATES 


161 


The  curve  may  be  plotted  by  solving  the  equation  for  p  and 
finding  the  values  of  p  for  particular  values  of  6  until  the  coor- 
dinates of  enough  points  are  obtained  to  determine  the  form  of 
the  curve. 

The  plotting  is  facilitated  by  the  use  of  polar  coordinate  paper,  which  enables 
us  to  plot  values  of  6  by  lines  drawn  through  the  pole  and  values  of  p  by  circles 
having  the  pole  as  center.  The  tables  on  p.  21  are  to  be  used  in  constructing 
tables  of  values  of  p  and  0. 

In  discussing  the  locus  of  an  equation  the  following  points 

should  be  noticed. 

1.  The  intercepts  on  the  polar  axis  are  obtained  by  setting 

^  =  0  and  6  =  TT  and  solving  for  p. 

But  other  values  of  9  may  make  p  =  0  and  hence  give  a  point  on  the  polar  axis, 
namely,  the  pole. 

2.  The  curve  is  symmetrical  with  respect  to  the  pole  if,  when 
—  p  is  substituted  for  p,  only  the  form  of  the  equation  is  changed. 

3.  The  curve  is  symmetrical  with  respect  to  the  polar  axis  if, 
when  —  ^  is  substituted  for  0,  only  the  form  of  the  equation  is 
changed. 

4.  The  directions  from  the  pole  in  which  the  curve  recedes  to 
infinity,  if   any,    are   found   by 
obtaining  those  values  of  0  for 
which  p  becomes  infinite. 

5.  The  method  of  finding  the 
values  of  $  which  must  be  ex- 
cluded, if  any,  depends  on  the 
given  equation. 

Ex.  1.  Discuss»and  plot  the  locus 
of  the  equation  p  =  10  cos  0. 

Solution.  The  discussion  enables  us 
to  simplify  the  plotting  and  is  there- 
fore put  first. 

1.  For  ^  =  0  p  =  10,  and  foY0  =  7t 
p  =  —  10.  Hence  the  curve  crosses  the 
polar  axis  10  units  to  the  right  of  the 
pole. 

2.  The  curve  is  symmetrical  with  respect  to  the  polar  axis,  for 
cos(-  0)  =  COS0  (4,  p.  19). 


0 

p 

d 

9 

0 

10 

Tt 

2 

0 

7t 

l2 

9.7 

nit 

12 

-  2.6 

7f 

6 

8.7 

2;r 
3 

-  5 

7t 

37r 

-  7 

4 

7 

4 

7t 

5 

hit 
6 

-  8.7 

3 

llTT 

-  9.7 

57t 

12 

12 

2.6 

It 

-10 

152 


ANALYTIC  GEOMETRY 


3.  As  cos  d  is  never  infinite, 
the  curve  does  not  recede  to 
infinity.  Hence  the  curve  is  a 
closed  curve. 

4.  No  values  of  6  make  p 
imaginary. 

Computing  a  table  of  values 
we  obtain  the  table  on  p.  151. 
As  the  curve  is  symmetrical 
with  respect  to  the  polar  axis, 
the  rest  of  the  curve  may  be 
easily  constructed  without  com- 
puting the  table  farther;  but 
as  the  curve  we  have  already 
constructed  is  symmetrical 
with  respect  to  the  polar  axis, 
no  new  points  are  obtained.     The  locus  is  a  circle. 

Ex.  2.    Discuss  and  plot  the  locus  of  the  equation  p"^  =  aP-  cos  2  0. 
Solution.  The  discussion  gives 
us  the  following  properties. 

1.  For  ^  =  0  or  7t  p  =  ±a. 
Hence  the  curve  crosses  the 
polar  axis  a  units  to  the  right 
and  left  of  the  pole. 

2.  The  curve  is  symmet- 
rical with  respect  to  the  pole. 

3.  It  is  also  symmetrical 
with  respect  to  the  polar 
axis,  for  cos  (—  2  ^)  =  cos 2  6 
(4,  p.  19). 

4.  p  does  not  become  infinite. 

5.  p  is  imaginary  when 
cos  2  ^  is  negative,  cos  2  d 
is  negative   when   2  ^  is    in 

the  second  or  third  quadrant;  that  is,  when 

i^>2»>*  or  I^>2»>^. 
2  2  2  2 

Hence  we  must  exclude  values  of  0  such  that 

^^^  n^  ^      J  77r     ^     57r 

—->e>-  and  —  >d> 

4  4  4  4 

The  accompanying  table  of  values  is  all  that 


e 

P 

e 

p 

0 
12 

±a 
±.93  a 

It 
6 
It 
4 

±.7a 
0 

need  be  computed  when  we  take  account  of  2,  3,  and  5. 


POLAR  COORDINATES 


153 


The  complete  curve  is  obtained  by  plotting  these  points  and  the  points 
symmetrical  to  them  with  respect  to  the  polar  axis.  The  curve  is  called  a 
lemniscate.     In  the  figure  a  is  taken  equal  to  9.5. 

Ex.  3.    Discuss  and  plot  the  locus  of  the  equation 

2 
^  ""  1  +  cos  ^  ' 
Solution.    1.  For  ^  =  0  p  =  1,  and  iov  d  =  it  p 
the  polar  axis  one  unit  to  the  right  of  the  pole. 

2.  The  curve  is  not  symmetrical  with  respect  to  the  pole, 
be  inferred  from  1  ? 

3.  The  curve  is  symmetrical  with  respect  to  the  polar  axis,  since 
cos(-  d)  =  cos^  (4,  p.   19). 

4.  p  becomes  infinite  when  1  +  cos  ^  =  0  or  cos  6  =—  \  and  hence 
6  =  7t.     The  curve  recedes  to  infinity  in  but  one  direction. 

5.  p  is  never  imaginary. 

On  account  of  3  the  table  of  values  is  computed  only  to  ^  =  tt,  and  the 
rest  of  the  curve  is  obtained  from  the  symmetry  with  respect  to  the  polar 

axis.     The  locus  is  a  parabola. 


CO  ;  so  the  curve  crosses 


How  may  this 


e 

p 

Q 

? 

0 

1 

lit 

2.7 

IT 

1.02 

12 

12 

7t 

1.07 

27r 
3 

4 

6 

7t 

4 

1.2 

3  7r 
4 

6.7 

TC 

1.3 

hit 

14 

3 

G 

57r 
12 

1.6 

ll;r 

50 

* 

12 

Tt 

2 

2 

It 

CO 

PROBLEMS 

Discuss  and  plot  the  loci  of  the  following  equations. 
1.  p  =  10.     e  =  tan-il.  5.  psin^  =  4. 


2.  /t)  =  5.     Q=- 

Z.  p  =  16.C0S  d. 
4.  p  COS  ^  =  6. 


6 


6.  p 


7.  p  = 


COS^ 


COS  6 


154 


ANALYTIC  GEOMETRY 


1  -  2  cos  ^ 
9.  p  =  a  sin  6. 

10.  p  =  a{l  —  cos^). 

11.  /)2  sin  2  ^  n:  16. 

12.  p2  =  16sin2^. 

13.  p^cos^2d  =  a^. 

14.  p  =  asm2  6.     p  =  acos2d. 
8 


15.  p  = 


1  —  e  cos  0 
for  e  =  1,  2,  i. 


16.  pcose  =  asin^d. 

17.  /9C0S  ^  =  a  cos  2d. 

18.  /)  =  a(4  +  6cos^) 

for  6  =  3,  4,  6. 

19.  p  = 

1  +  tan  ^ 

20.  p  =  asece  ±h 

for  a > 6,  a  =  6,  a<b. 

21.  p  =  ad. 

22.p  =  asmS6.     p  =  acosSd. 


23.  Prove  that  the  locus  of  an  equation  is  symmetrical  with  respect  to 

e  =  —  it  the  results  of  substituting  -  +  d  and  —-  0  give  equations  which 

2  2  ^ 

differ  only  in  form. 

24.  Apply  the  test  for  symmetry  in  problem  23  to  the  loci  of  4,  5,  10,  11, 
and  12. 

62.  Transformation  from  rectangular  to   polar  coordinates. 

Let  OX  and  OF  be  the  axes  of  a  rectangular  system  of  coordi- 
nates, and  let  0  be 
the  pole  and  OX  the 
polar  axis  of  a  sys- 
tem of  polar  coor- 
dinates. Let  (x,  y) 
and  (p,  6)  be  respec- 
tively the  rectan- 
gular and  polar  coor- 
dinates of  any  point 
P.     It  is  necessary 

to  distinguish  two  cases  according  as  p  is  positive  or  negative. 
When  p  is  positive  (Fig.  1)  we  have,  by  definition, 


cos  6 

whatever  quadrant  P  is  in. 
Hence 


sin  6 


(1) 


X  =  p  cos  0,  y  =  p  sin  B. 


POLAR  COORDINATES  155 

When  p  is  negative  (Fig.  2)  we  consider  the  point  P'  symmet- 
rical to  P  with  respect  to  0,  whose  rectangular  and  polar  coordi- 
nates are  respectively  (—  x,  —  y)  and  (—  p,  6).  The  radius  vector 
of  P',  —  p,  is  positive  since  p  is  negative,  and  we  can  therefore  use 
equations  (1).     Hence  for  P' 

—  £c  =  —  p  cos  ^,  —  y  =  —  p^md\ 
and  hence  for  P 

X  =      p  cos  $,      y  =  p  sin  df 
as  before. 

Hence  we  have 

Theorem  I.  If  the  pole  coincides  with  the  origin  and  the  polar 
axis  with  the  positive  X-axis,  then 

a?  =  yocos  9, 
p  sin  $, 

where  (x,  y)  are  the  rectangular  coordinates  and  (p,  6)  the  polar 
coordinates  of  any  point. 

Equations  I  are  called  the  equations  of  transformation  from  rec- 
tangular to  polar  coordinates.  They  express  the  rectangular 
coordinates  of  any  point  in  terms  of  the  polar  coordinates  of 
that  point  and  enable  us  to  find  the  equation  of  a  curve  in  polar 
coordinates  when  its  equation  in  rectangular  coordinates  is  known, 
and  vice  versa. 

From  the  figures  we  also  have 


<■)  {: 


(2) 


p^  =  00^  -{■  y^,  0  =  tan-i  ^, 


oc 


e  =  — ^  cos^=       ^ 


±  Vx2  -^y^  ±  Vic2  +  y2 

These  equations  express  the  polar  coordinates  of  any  point  in  terms 
of  the  rectangular  coordinates.  They  are  not  as  convenient  for  use 
as  (I),  although  the  first  one  is  at  times  very  convenient. 

Ex.  1.    Find  the  equation  of  the  circle  x'^  +  y'^  =  25  in  polar  coordinates. 

Solution.  Substitute  the  values  of  x  and  ?/  given  by  (I).  This  gives 
p2  cos2  e  +  p^  sin2  d  =  25,  or  (by  3,  p.  19)  p^  =  2b;  and  hence  p  =  ±  5,  which 
is  the  required  equation.  It  expresses  the  fact  that  the  point  {p,  0)  is  five 
units  from  the  origin. 


156  ANALYTIC  GEOMETRY 

Ex.  2.    Find  the  equation  of  the  lemniscate  (Ex.  2,  p.  152)  p^  =  a^  cos  2  6 
in  rectangular  coordinates. 

Solution.    By  14,  p.  20,  we  have 

/)2=  a2(cos2^-sin2^). 
Multiplying  by  p%  p^  =  a^  (^2  cos2  d  -  p"^  sin2  d). 

From  (2)  and  (I),  {x^  +  2/2)2  =  a'^^x2  _  ^2),    ^^g. 

63.  Applications. 

Theorem  II.    The  general  equation  of  the  straight  line  in  polar 
coordi?iates  is 

(II)  p(A  cos  6 -hB  sin  e)  +  C  =  0, 
where  A,  B,  and  C  are  arbitrary  constants. 

Proof.    The  general  equation  of  the  line  in  rectangular  coordi- 
nates is  (Theorem  II,  p.  S(y) 

Ax  +  By  +  C  =  0.  }j) 

By  substitution  from  (I)  we  o.btain  (II).  q.e.d. 

When  A  =  0  the  line  is  parallel  to  the  polar  axis,  when  7?  =  0  it  is  perpen- 
dicular to  the  polar  axis,  and  when  C  =  0  it  passes  through  the  pole. 

In  like  manner  we  obtain 

Theorem  III.    The  general  equation  of  the  circle  in  polar  coordi- 
nates is 

(III)  p2  ^  p  (i)  cos  |9  +  J^  sin  e)J^F  =  0, 
where  D,  E,  and  F  are  arbitrary  constants. 

Corollary.    If  the  pole  is  on  the  circumference  and  the  polar  axis 
passes  th7vugh  the  center,  the  equation  is 
p  —  2  r  cos  ^  =  0, 
where  r  is  the  radius  of  the  circle. 

For  if  the  center  lies  on  the  polar  axis,  or  X-axis,  E=0  (Corollary,  p.  131) ; 
and  if  the  circle  passes  through  the  x)ole,  or  origin,  F=0.    The  abscissa  of  the 

D 
center  equals  the  radius,  and  hence  (Theorem  I,  p.  131)  —  —  =  r,  or  I)  —  —  2r. 

Substituting  these  values  of  D,  E,  and  F  in  (III)  gives  /o  —  2r  cos  ^  =  0. 

Theorem  IV.    The  length  I  of  the  line  joining  two  points  P^  (pi,  Oi) 
and  P2  (p2,  O2)  is  given  by 

(lY)  P  =  p,^  +  /),^  -  2  p,p,  cos  {d,  -  9,). 


POLAR  COORDINATES  157 

Proof.    Let  the  rectangular  coordinates  of  P^  and  Pg  be  respec- 
tively (iCj,  ?/i)  and  (x2,  y^.     Then  by  Theorem  I,  p.  155, 

Xi  =  pi  cos  6if  X2  =  p2  cos  62, 

2/1  =  pi  sin  $1,  1/2  =  p2  sin  0^. 
By  Theorem  IV,  p.  31, 

l'  =  {x,-X2Y-\-{y,-y2y, 
and  hence       l^  =  (p^  cos  61  —  pa  cos  ^2)^  +  (pi  sin  61  —  pa  sin  $2)^. 

Removing  parentheses  and  using  3,  p.  19,  and  11,  p.  20,  we 
obtain  (lY).  <^  -^^^         q.e.d. 

PROBLEMS 

1.  Transform  the  following  equations  into  polar  coordinates  and  plot 
their  loci. 

(a)  x-Sy  =  0.  Ans.    6  =  tan-'  (i). 

K 

(b)  ?/ +  5  =  0.  Ans.   p  = 

sin^ 

(c)  x^-{-y^  =  16.  Ans.  p=±4. 

(d)  x^  +  y'^  —  ax  =  0.  Ans.  p  =  acosd. 

(e)  2xy  =  7.  Ans.  p^ sin 2 ^  =  7. 

(f)  x-^  -  ?/2  =  a2.  Ans.  p^  cos  2  ^  =  a^. 

(g)  xcosu  +  ysinu  —  p  =  0.  ^ns.  pcos  {d  —  oj)  ~ p  =  0. 

(h)  (1  -  e2)  x2  +  ?/  -  2  e2]9x  -  e2p2  =  0.  ^ns.    p  =  — -^ 

1  —  ecos^ 

(i)  2a:?/  +  4?/2-8a;  +  9  =  0.    Ans.  p2(sin2^  +  4sin2^)  -  8/)COs6' +  9  =  0. 

2.  Transform  equations  1  to  21,  p.  153,  into  rectangular  coordinates. 

3.  Find  the  polar  coordinates  of  the  points  (3,  4),  (—4,  3),  (5,  —12), 
(4,  5). 

4.  Find  the  rectangular  coordinates  of  the  points  (5,  —  ),  (  —  2,  —  ), 
,0   ^x  '  \      2/     \  4  / 

ep 


5.  Transform  into  rectangular  coordinates  p  = 


ecos^ 


64.  Equation  of  a  locus.  The  equation  of  a  locus  may  often 
be  found  with  more  ease  in  polar  than  in  rectangular  coordinates, 
especially  if  the  locus  is  described  by  the  end  of  a  line  of  vai'iable 
length  revolving  about  a  fixed  point.  The  steps  in  the  process 
of  finding  the  polar  equation  of  a  locus  correspond  to  those  in  the 
Rule  on  p.  53. 


158 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  locus  of  the  middle  points  of  the  chords  of  the  circle 
C:  p  —  2r  cos ^  =  0  which  pass  through  the  pole  which  is  on  the  circle. 

Solution.  Let  P{p,  6)  be  any  point  on  the  locus.     Then,  by  hypothesis, 

OP  =  ^  OQ, 

where  Q  is  a  point  on  C. 

But      OP  =  /)  and  OQ  =  2  r  cos  6. 

Hence  p  =  r  cos  6. 

From  the  Corollary  (p.  156)  it  is  seen  that 
the  locus  is  a  circle  described  on  the  radius 
of  C  through  0  as  a  diameter. 

Ex.  2.    The  radius  of  a  circle  is  prolonged  a  distance  equal  to  the  ordinate 
of  its  extremity.     Find  the  locus  of  the  end  of  this  line. 

Solution.    Let  r  be  the- radius  of  the  circle,  let  its  center  be  the  pole,  and 
let  P  (p,  6)  be  any  point  on  the  locus.     Then, 
by  hypothesis, 

OP  =  OjB  +  CB. 

But  OP  =  p, 

OB  =  r, 

and  CB  =  r  sin  6. 

Hence  the  equation  of  the  locus  of  P  is 

p  =  r  +  >•  sin  ^. 
The  locus  of  this  equation  is  called  a  cardioid. 


p{P,d) 


PROBLEMS 

1.  Chords  passing  through  a  fixed  point  on  a  circle  are  extended  their 
own  lengths.     Find  the  locus  of  their  extremities. 

Ans.    A  circle  whose  radius  is  a  diameter  of  the  given  circle. 

2.  Chords  of  the  circle  p  =  10  cos  6  which  pass  through  the  pole  are 
extended  10  units.     Find  the  locus  of  the  extremities  of  these  lines. 

Ans.   p  =  10  (1  +  cos  d). 

3.  Chords  of  the  circle  p  =  2a cos 6  which  pass  through  the  pole  are 
extended  a  distance  2  6.    Find  the  locus  of  their  extremities. 

Ans.  p  =  2  (6  +  a  cos  6).  , 


POLAR  COORDINATES  169 

4.  Find  the  locus  of  the  middle  points  of  the  lines  drawn  from  a  fixed 
point  to  a  given  circle. 

Hint.  Take  the  fixed  point  for  the  pole  and  let  the  polar  axis  pass  through  the  center 
of  the  circle. 

Ans.   A  circle  whose  radius  is  half  that  of  the  given  circle  and  whose 
center  is  midway  between  the  pole  and  the  center  of  the  given  circle. 

6.  A  line  is  drawn  from  a  fixed  point  0  meeting  a  fixed  line  in  Pi.     Find 
the  locus  of  a  point  P  on  this  line  such  that  OPi  •  OP  =  a^.   Ans.   A  circle. 

6.  A  line  is  drawn  through  a  fixed  point  0  meeting  a  fixed  circle  in  Pi 
and  Pa.     Find  the  locus  of  a  point  P  on  this  line  such  that 

OP  =  2  ^^^ '  ^^^  .       Ans.   A  straight  line. 
OPi  +  OP2 


CHAPTER  VII 
TRANSFORMATION  OF  COORDINATES 

65.  When  we  are  at  liberty  to  choose  the  axes  as  we  please 
we  generally  choose  them  so  that  our  results  shall  have  the  sim- 
plest possible  form.  When  the  axes  are  given  it  is  important 
that  we  be  able  to  find  the  equation  of  a  given  curve  referred  to 
some  other  axes.  The  operation  of  changing  from  one  pair  of 
axes  to  a  second  pair  is  known  as  a  transformation  of  coordinates. 
We  regard  the  axes  as  moved  from  their  given  position  to  a  new 
position  and  we  seek  formulas  which  express  the  old  coordinates 
in  terms  of  the  new  coordinates. 


66.  Translation  of  the  axes.    If  the  axes  be  moved  from" a  first 

position  OX  and  OF  to  a  second  position  O'Z'  and  O'F'  such  that 

O'X'  and  O'F'  are  respectively  parallel  to  OX  and  OY,  then  the 

axes  are  said  to  be  translated  from  the  first  to  the  second  position. 

Let  the  new  origin  be  0\h,  k)   and  let  the    coordinates  of 

any  point  P  before  and  after  the 
translation  be  respectively  (x,  y) 
and  (x\  ?/').  Projecting  OP  and 
OO'P  on  OX,  we  obtain  (Theorem 
XI,  p.  48) 

X  =  x^  -\-  h. 

Similarly,  y  =  y'  -{-  k. 

Hence, 

Theorem  I.  If  the  axes  be  translated  to  a  netv  origin  (h,  k),  and 
if  (x,  y)  and  (x',  y')  are  respectively  the  coordinates  of  any  point  P 
before  and  after  the  translation,  then 


Y' 

Y' 

' 

■  N 

1 

d 

^^^ 

B 

(h. 

ky 

X         Y 

X 

0 

/" 

/ 

» 

A           M 

X 

(I) 


•£c  =  a?'  +  h, 

y  =  y^  ■\-k. 

160 


TRANSFORMATION  OF  COORDINATES 


161 


Equations  (I)  are  called  the  equations  for  translating  the  axes. 
To  find  the  equation  of  a  curve  referred  to  the  new  axes  when 
its  equation  referred  to  the  old  axes  is  given,  we  substitute  the 
values  of  x  and  y  given  by  (I)  in  the  given  equation.  For  the 
given  equation  expresses  the  fact  that  P  (a?,  y)  lies  on  the  given 
curve,  and  since  equations  (I)  are  true  for  all  values  of  (a:,  y),  the 
new  equation  gives  a  relation  between  x'  and  y^  which  expresses 
that  P{x\  2/')  lies  on  the  curve  and  is  therefore  (p.  53)  the  eqiiar 
tion  of  the  curve  in  the  new  coordinates. 

Ex.  1.    Transform  the  equation 

x2  +  y2_6x  +  4y-12  =  0 
when  the  axes  are  translated  to  the  new  origin  (3,  —  2). 

Solution.    Here  ^  =  3  and  k=—2,  so  equations  (1)  become 

X  =  x'  -\-  S,  y  =  y'  -  2. 
Substituting  in  the  given  equation,  we 
obtain 
(x'  +  3)2  +  (z/'-2)2-6(x'  +  3) 

+  4(^-2) -12  =  0, 

or,  reducing,    x'^  +  y'^  =  25. 

This  result  could  easily  be  foreseen. 
For  the  locus  of  the  given  equation  is 
(Theorem  I,  p.  131)  a  circle  whose  center  is 
(3,  -2)  and  whose  radius  is  5.  When  the 
origin  is  translated  to  the  center  the  equa- 
tion of  the  circle  must  necessarily  have 
the  form  obtained  (Corollary,  p.  58). 

PROBLEMS 

1.  Find  the  new  coordinates  of  the  points  (3,  -  5)  and  (-  4,  2)  when  the 
axes  are  translated  to  the  new  origin  (3,  6). 

2.  Transform  the  following  equations  when  the  axes  are  translated  to  the 
new  origin  indicated  and  plot  both  pairs  of  axes  and  the  curve. 

Ans.   Sx'  -4y'  =  0. 


Y 

£ 

" 

^ 

-^ 

^ 

■^ 

N 

' 

/ 

\ 

\ 

0 

\ 

X 

0' 

(3, 

2) 

X' 

/ 

/ 

\ 

/ 

V 

^ 

y 

Ans.     iC'2  +  2/'2  ^ 

Ans.   y"^  =  6  x'. 


(a)  3x-4y  =  6,  (2,0). 

(b)  x2  +  2/2  -  4iC  -  2 ?/  =  0,  (2,  1). 

(c)  y2-6x  +  9  =  0,  (1,0). 

(d)  x2  +  2/2_i  =  o,  (-3,  -2). 

(e)  ?/2_2fcx  +  fc2  =  0,  (-,  oV 

(f)  x2-4?/2+8x+24?/-20=0,  (-4,  3).   Ans.    x'2-4i/'2  =  0 


Ans. 
Ans. 


x'2+?/'2_6a;'_4y'+12=0. 
y'2  -  2  kx'. 


162 


ANALYTIC  GEOMETRY 


3.  Derive  equations  (I)  if  0'  is  in  (a)  the  second  quadrant ;  (b)  the  third 
quadrant ;   (c)  the  fourth  quadrant. 

67.  Rotation  of  the  axes.  Let  the  axes  OX  and  0  F  be  rotated 
about  0  through  an  angle  6  to  the  positions  OX'  and  0Y\  The 
equations  giving  the  coordinates  of  any  point  referred  to  OX  and 
OF  in  terms  of  its  coordinates  referred  to  OX'  and  OY'  are  called 
the  equations  for  rotating  the  axes. 

Theorem  II.    The  equations  for  rotating  the  axes  through  an  angle 


6  are 


(II) 


05  =  a?'  cos  5  —  y^  sin  9y 
2/  =  0?'  sin  ^  +  y^  cos  9, 


Proof.    Let  P  be  any  point 
whose  old   and    new   coordi- 

^^^ ^   nates  are  respectively  (x,  y) 

^     and   (x',  y').     Draw  OP  and 
draw  PM'  perpendicular  to  OX'.     Project  OP  and  OM'P  on  OX. 


The  proj.  of  OP  on  OX  =  x. 

The  proj.  of  OM'  on  OX  =  x'  cos  6. 


(Theorem  III,  p.  31) 
(Theorem  II,  p.  30) 

The  proj.  of  M'P  on  OX  =  y'Gos['^  +  e\   (Theorem  II,  p.  30) 


=  —  y  sm  &. 

Hence  (Theorem  XI,  p.  48) 

X  ==  x'  cos  0  —  y'  sin  6. 


(by  6,  p.  20) 


In  like  manner,  projecting  OP  and  OM'P  on  OF,  we  obtain 


y  =z  x'  COS 


0\-\-y'cosO 


x'  sin  0  -\-  y'  cos  0. 


Q.E.D. 


If  the  equation  of  a  curve  in  x  and  y  is  given,  we  substitute 
from  (II)  in  order  to  find  the  equation  of  the  same  curve  referred 
to  OX'  and  OY'. 


TRANSFORMATION  OF  COORDINATES 


163 


Ex.  1.    Transform  the  equation  x^  —  y^  =  16  when  the  axes  are  rotated 
It 


through 


Solution.    Since 


.     7t      1    /-        1 
sm  -  =  -  V 2  =  -— 
4      2  V2 

and      cos  — =  — -i 
4      V2 

equations  (II)  become 

x'  -  y'  x'  +  y' 


Vi 


V^ 


Substituting    in    tlie     given 
equation,  we  obtain 


Wttf 

\  s 

yirt- 

:  !s  s^ 

Z     1     " 

\  s 

Z     / 

-    -^    s 

Z     t     ~- 

-     4^     5 

41            ^ 

X  ^ 

Js                        -r 

:    7    z 

s^  t     " 

-  -7  z 

s  _s 

-  7-  z 

s^^  I 

-/-A      - 

( ^'  -y'  V  _  /  ^'  -^y'  Y 


or,  simplifying, 


x^  +  y'V 
V2   ^      ^    V2 

x'?/'  +  8  =  0. 


16, 


PROBLEMS 

1.  Find  the  coordinates  of  the  points  (3,  1),  (—2,  6),  and  (4,  —  1)  when 

It 
the  axes  are  rotated  through  —  • 

2.  Transform  the  following  equations  when  the  axes  are  rotated  through 
the  indicated  angle.     Plot  both  pairs  of  axes  and  the  curve. 

Tt 


(a)  X  -  y  =  0, 

(b)  x2  +  2xy  +  2/2  =  8,  ^. 

4 

(c)  2/2  =  4x,  -|. 

(d)  X2  +  4X2/  +  2/2  3,16,^. 

4 

(e)  x2  +  2/2  =  r2,  e. 

(f)  x2  +  2  x?/  +  2/2  +  4 X  -  4  2/  =  0,  -  -. 

4 


Ans.  y'  =  0. 

J.ns.  x'2  =  4. 

-4  ns.  x'2  =  4  2/'. 

^ns.  3x'2-2/'2  =  16. 

Ans.  x'2  +  2/'2  =  r2. 


3.  Derive  equations  (II)  if  6  is  obtuse. 

68.  General  transformation  of  coordinates.  If  the  axes  are 
moved  in  any  manner,  they  may  be  brought  from  the  old  position 
to  the  new  position  by  translating  them  to  the  new  origin  and 
then  rotating  them  through  the  proper  angle. 


164 


ANALYTIC  GEOMETRY 


^j-j-j.  (i€  =  o(^'  COS  O-y^ 

^      ^  12/  =  a?'  sin  9  +  y' 


Y 

^'\ 

^' 

X" 

0 

/^ 

v^ 

^' 

Theorem  III.  If  the  axes  he  translated  to  a  new  origin  (Ji,  k)  and 
then  rotated  through  an  angle  0,  the  equations  of  the  transforma- 
tion of  coordinates  are 

oe'  cos  $  —  y'  &in  6  -\-  7i, 
cos  6  +  7c. 

Proof    To  translate  the  axes  to  O'X"  and  O'Y"  we  have,  by  (I), 

x  =  x"  -\-  h, 

where  (x",  t/")  are  the  coordi- 
nates of  any  point  P  referred 
to  O'X"  and  O'Y". 

To  rotate  the  axes  we  set, 
by  (II),       , 

x"  =  x'  cos  6  —  y'  sin  6, 
y"  =  x'  sin  0  -\-  y'  cos  0. 
Substituting  these  values  of  x"  and  y",  we  obtain  (III).      q.e.d. 

69.  Classification  of  loci.  The  loci  of  algebraic  equations 
(p.  17)  are  classified  according  to  the  degree  of  the  .equations. 
This  classification  is  justified  by  the  following  theorem,  which 
shows  that  the  degree  of  the  equation  of  a  locus  is  the  same  no 
matter  how  the  axes  are  chosen. 

Theorem  IV.  The  degree  of  the  equation  of  a  locus  is  unchanged 
by  a  transformation  of  coordinates. 

Froof  Since  equations  (III)  are  of  the  first  degree  in  x'  and 
y',  the  degree  of  an  equation  cannot  be  raised  when  the  values  of 
X  and  y  given  by  (III)  are  substituted.  Neither  can  the  degree 
be  lowered;  for  then  the  degree  must  be  raised  if  we  transform 
back  ta  the  old  axes,  and  we  have  seen  that  it  cannot  be  raised 
by  changing  the  axes.^ 

As  the  degree  can  neither  be  raised  nor  lowered  by  a  trans- 
formation of  coordinates,  it  must  remain  unchanged.  q.e.d. 


*  This  also  follows  from  the  fact  that  when  equations  (III)  are  solved  for  x'  and  ; 
results  are  of  the  first  degree  in  x  and  y. 


the 


TRANSFORMATION  OF  COORDINATES  165 

70.  Simplification  of  equations  by  transformation  of  coordi- 
nates. The  principal  use  made  of  transformation  of  coordinates 
is  to  discuss  the  various  forms  in  which  the  equation  of  a  curve 
may  be  put.  In  particular,  they  enable  us  to  deduce  simjde  forms 
to  which  an  equation  may  be  reduced. 

Rule  to  simplify  the  form  of  an  equation. 

First  step.  Substitute  the  values  of  x  and  y  given  by  (I)  [or  (II)] 
and  collect  like  powers  of  x'  and  y'. 

Second  step.  Set  equal  to  zero  the  coefficients  of  two  ter'ms 
obtained  in  the  first  step  which  contain  h  and  k  (or  one  coeffi- 
"cient  containing  6). 

Third  step.  Solve  the  equations  obtained  in  the  seco7id  step  for 
h  and  k*  (or  9). 

Fourth  step.  Substitute  these  values  for  h  and  k  (or  &)  in  the 
result  of  the  first  step.     The  result  will  be  the  required  equation. 

In  many  examples  it  is  necessary  to  apply  the  rule  twice  in 
order  to  rotate  the  axes,  and  then  translate  them,  or  vice  versa. 
It  is  usually  simpler  to  do  this  than  to  employ  equations  (III) 
in  the  Rule  and  do  both  together.  Just  what  coefficients  are 
set  equal  to  zero  in  the  second  step  will  depend  on  the  object 
in  view. 

It  is  often  convenient  to  drop  the  primes  in  the  new  equation 
and  remember  that  the  equation  is  referred  to  the  new  axes. 

Ex.  1.  SimpUfy  the  equation  y2  _  gx  +  6 y  +  17  =  0  by  translating  the 
axes. 

Solution.    First  step.    Set  x  =  x''  -\-  h  and  y  =  y'  -\-  k. 

This  gives  {y'  +  A;)2  -  8  (x'  +  7i)  +  6  (?/'  +  A:)  +  17  =  0,  or 

(1)  y'2_8x'  +  2A:    ?/'+     A:2   f  =  0. 

+  6 


y'+    k^ 

-8h 

+  6A: 

+  17 

*  It  may  not  be  possible  to  solve  these  equations  (Theorem  IV,  p.  90). 

t  These  vortical  bars  play  the  part  of  parentheses.  Thus  2  i-  +  6  is  the  coefficient  of  y' 
and  k^-8h  +  Gk+  17  is  the  constant  term.  Their  use  enables  us  to  collect  like  powers 
of  x^  and  y'  at -the  same  time  that  we  remove  the  parentheses  in  the  preceding  equation. 


166 


ANALYTIC  GEOMETRY 


Second  step.    Setting  the  coefficient  of  y'  and  the  constant  term,  the  only 
,  coefficients  containing  h  and  A;,  equal  to  zero,  we 

obtain 

(2)  2  fc  +  6  =:  0, 

(3)  A;2- 8/1 +  6  A; +  17  =  0. 

Third  step.  Solving  (2)  and  (3)  for  h  and  fc, 
we  find 

A;  =-3,  /i=:l. 

Fourth  step.    Substituting  in  (1),  remember- 
ing that  h  and  Ic  satisfy  (2)  and  (3),  we  have 
y'l  _  8  X'  =  0. 

The  locus  is  the  parabola  plotted  in  the  figure 
which  shows  the  new  and  old  axes. 

Ex.  2.  Simplify  x2  +  4?/2-2x-16?/  +  l=0 
by  translating  the  axes. 

Solution.  First  step.  Set  x  =x'  +  h,  y=y' +k. 
This  gives 


Y' 

^r- 

^ 

/ 

/ 

/ 

0 

X 

s 

(i 

-3) 

X' 

\ 

\ 

s 

s 

S 

(4) 


x"^+^y"^-\-2h 


X'  +  ^k 

y'+    h^ 

-.16 

+  4A:2 

~2h 

-16k 

+  1 

=  0. 


Second  step.    Set  the  coefficients  of  x'  and  y'  equal  to  zero.     This  gives 

2h-2  =  0,  Sk-16  =  0. 
Third  step.    Solving,  we  obtain 
^1  =  1,  A:  =  2. 

Fourth  step .    Substituting  in  (4) ,  we  obtain 
x'2  +  4  y'2  ^  16. 

Plotting  on  the  new  axes,  we  obtain  the 
figure. 

Ex.  3.    Remove  the  xy-term  from  x'^  -}-  ixy  +  y^  =  4hj  rotating  the  axes. 
Solution.    First  step.    Set  x  =  x'  cos  ^  —  y'  sin  6  and  y  =  x'sme  +  y'  cos  ^, 
whence 


+  4  sin  0  cos  e 
+  sin2  e 


?/2  =  4. 


x'2  -  2  sin  ^  cos  d 

+  4  (cos2  d  -  sin2  d) 

-{- -2  &m  d  cos,  6 
or,  by  3,  p.  19,  and  14,  p.  20, 
(5)  (1  +  2sin2^)x'2  +  4cos2^.xV+(l  -  2  sin2^)y'2  ^  4. 


x'y'-\-  sin2  0 
— 4sin^cos^ 
4-  cos2  6 


TRANSFORMATION  OF  COORDINATES 


167 


Second  step.    Setting  the  coefficient  of  xfy'  equal  to  zero,  we  have 

cos  2  ^  =  0. 
Third  step.    Hence 


2^  = 


->[  1 1 1  \\ 

itrtttP 

-    S         A 

z  . 

-7-     - 

S            ^ 

z 

S       ^ 

^-Z         - 

"^^                 S 

^Z3I 

^^v^      j\ 

^^2 

-              ^ 

^    ""     ? 

7\ 

~Z 

s:  s        : 

7 

4     s^ 

z 

^i_    s 

'  z 

V       ^   - 

'7 

A 

INNlK 

Fourth  step.  Substituting  in 
(5),  we  obtain,  since  sin  —  =  1 
(p.  21), 

The  locus  of  this  equation  is 
the  hyperbola  plotted  on  the 
new  axes  in  the  figure. 


'     From  cos 2  ^  =  0  we  get,  in  general,  26  =  — \-  nit,  where  n  is  any  positive 

7t  Tt 

or  negative  integer,  or  zero,  and  hence  6  =  —  +  n  —  -     Then  the  xy-term  may 

4  2 

be  removed  by  giving  d  any  one  of  these  values.     For  most  purposes  we 
choose  the  smallest  positive  value  of  6  as  in  this  example. 

Ex.  4.    Simplify  x3  4-6x2  +  12x  —  4y  +  4  =  0  by  translating  the  axes. 

Solution.    First  step.    Set 

x  =  x'  -{-  h,  y  =  y'  +  k. 
We  obtain 

(6)  x'^-i-Sh   x'^+    3/i2   x'-4?/+      h^    =  0. 
+  6  -\-l2h  +    6/i2 

+  12  +  12  ;i 

-    ik 
+    4 
Second  step.    Set  equal  to  zero  the  coefficient 
of  x'2  and  the  constant  term.     This  gives 
3  /i  +  6  =  0, 
7^3  4.  6  ^2  +  12  /i  -  4  A;  -f  4  =  0. 
Third  step.    Solving, 

h  =  -2,  k  =  -l. 

Fourth  step.    Substituting  in  (6),  we  obtain 

x'3  _  4  ?/'  =  0, 


YA 

r| 

1 

' 

i 

^ 

yc 

.Y 

/ 

^' 

(-2 

rl) 

x' 

f 

1 

1 

1 

1 

^ 

— 

— 

— 



- 

whose  locus  is  the  cubical  parabola  in  the  figure. 


168 


ANALYTIC  GEOMETRY 


PROBLEMS 

1.  Simplify  the  following  equations  by  translating  the  axes.     Plot  both 
pairs  of  axes  and  the  curve. 

(a)  x2  +  6  X  +  8  rr  0. 

(b)  x^-4y  +  S  =  0. 

(c)  x2  +  2/2  +  4  X  -  6  ?/  -  3  =  0. 

(d)  2/2  -  6  X  -  10  ?/  +  19  =  0. 

(e)  x2  -  2/2  +  8x  -Uy  -  33  =  0. 

(f)  x2  +  42/2  -  16xV  24y  +  84  =  0. 

(g)  2/^  +  8x-40  =  0. 
(h)  x3  -  2/2  +  14  2/  -  49  =  0. 
(i)  4x2 -4x2/+ 2/2 -40x^2«^2/-|;  99  =  0. 

2.  Remove  the  xy-term  from  the  following  equations  by  rotating  the  axes. 
Plot  both  pairs  of  axes  and  the  curve. 

(a)  x2  -  2  xy  +  y^  =  12. 

(b)  x2  -  2  X2/  +  2/2  +  8  X  +  8  2/  =  0. 

(c)  xy  =  18. 

(d)  25  x2  +  14  X2/  +  25  y^  =  288. 

(e)  3x2 -10x2/  + 3  2/2  =  0. 

(f )  6  x2  +  20  V3  xy  +  26  y2  =  324. 


Ans. 

x'2  =  1. 

Ans. 

x'2  1=  4  2/'. 

Ans. 

x'2  +  2/'2  =  16. 

Ans. 

2/'2  =  6x'. 

Ans. 

x'2  _  ^'2  =  0. 

Ans. 

x'2  +  4  2/'2  =  16. 

Ans. 

8  x'  +  2/'"^  =  0. 

Ans. 

2/'2  =  X'3. 

Ans. 

(2x'- 2/0^-1=0 

^ns. 

2/'2  =  6. 

^ns. 

y/2y'^  +  8x'  =  0. 

^)IS. 

x'2  _  2/'2  =  36. 

^ws. 

16  x'2 +  92/^2=144 

Ans. 

x'2  _  4  2/'2  =  0. 

Ans. 

9x'2-2/'2  =  81. 

71.  Application  to  equations  of  the  first  and  second  degrees. 

In  this  section  we  shall  apply  the  Rule  of  the  preceding  section 
to  the  proof  of  some  general  theorems. 

Theorem  V.    Bi/  rtioving  the  axes  the  general  equation  of  the  first 

degree, 

Ax  +  By  +C  =  Oj 

may  he  transformed  into  x'  =  0. 

Proof    Apply  the  Rule  on  p.  165,  using  equations  (III). 
Set  X  —  x^  cos  0  —  y'  sin  6  -\-  h, 

^y  =  x'  sin  0  -{-  y'  cos  0  +-  k. 
This  gives 


(1) 


A  cos  6 
-\-BsmO 


x'  —  A  sin  0 

y'  +  Ah 

-^  Boose 

+  Bk 

+  C 

=  0. 


TRANSFORMATION  OF  COORDINATES 


169 


Setting  the  coefficient  of  ?/'  and  the  constant  term  equal  to  zero 
gives 

(2)  -  ^  sin  (9  +  j5  cos  ^  =  0, 

(3)  Ah  +  Bk-^C  =  0. 


From  (2), 


tan  ^  =  "T '    or 


tan" 


From  (3)  we  can  determine  many  pairs  of  values  of  h  and  /c. 
One  pair  is 

A 


\ 


Substituting  in  (1)  the  last  two  terms  drop  out,  and  dividing 
by  the  coefficient  of  x'  we  have  left  £c'  =  0.  q.e.d. 

We  have  moved  the  origin  to  a  point  (h,  k)  on  the  given  line 
L,  since  (3)  is  the  condition  that  {h,  k)  lies  on  the  line,  and  then 
rotated  the  axes  until  the  new  axis  of 
y  coincides    with    L.      The   particular 
point  chosen  for  (h,  k)  was  the  point  0' 
where  L  cuts  the  Z-axis. 

This  theorem  is  evident  geometric- 
ally. For  ic'  =  0  is  the  equation  of 
the  new  F-axis,  and  evidently  any  line 
may  be  chosen  as  the  F-axis.  But  the  theorem  may  be  used  to 
prove  that  the  locus  of  every  equation  of  the  first  degree  is  a 
straight  line,  if  we  prove  it  as  above,  for  it  is  evident  that  the 
locus  of  ic'  =  0  is  a  straight  line. 

Theorem  VI.    The  term  in  xy  may  always  he  removed  from  an 
equation  of  the  second  degree, 

Ax^  +  Bxy  -\-Cy^-^Dx+Ey-\-F=Oj 

by  rotating  the  axes  through  an  angle  6  such  that 


(VI) 


tan2^  = 


A-  C 


170 


ANALYTIC  GEOMETRY 


Proof.    Set 
and 

This  gives 
(4)  y1cos2<9 

-\-B  sin  ^cos  6 
+  C  sin2  e 


X  =  x'  COS  0  —  y'  sin  0 
y  =  x'  sin  0  -{-  y'  cos  ^. 


—  B  sin  ^  cos  ^ 
+  C  cos^  ^ 


a;'  —  Z)  sin  $ 
-{-  Ecosd 


x'^-2AsmeG0se 
+  B(Gos^O-sm^e) 
+  2Csiiie^ose 

+  Dcose 

■i-  EsinO 

Setting  the  coefficient  of  x'y'  equal  to  zero,  we  have 
(C  -A)2  sin  ecosO  +  B  (cos^  0  -  sin^  0)  =  0, 
or  (14,  p.  20),  (C  -  ^)  sin  2  ^  +  5  cos  2  ^  =  0. 

B 


y'-\-F=0. 


Hence 


tan2<9 


A  -C 

If  0  satisfies  this  relation,  on  substituting  in  (4)  we  obtain  an 
equation  without  the  term  in  xy.  q.e.d. 

Corollary.  In  transforming  an  equation  of  the  second  degree  hy 
rotating  the  axes  the  constant  term  is  unchanged  unless  the  new 
equation  is  multiplied  or  divided  hy  some  constant. 

For  the  constant  terra  in  (4)  is  the  same  as  that  of  the  given  equation. 

Theorem  VII.  The  terms  of  the  first  degree  m,ay  he  removed  from, 
an  equation  of  the  second  degree, 

Ax^  H-  Bxy  +  C//2  +  Dx  +  Ey  -\-  F  =  0, 
hy  translating  the  axes,  provided  that  the  discriminant  of  the  terms 
of  the  second  degree,  A  =  ^^  —  4  yl  C,  is  not  zero. 

Proof     Set  X  =  x'  -\-  h,  y  =  y'  -\-  k. 

This  gives 


(5)      Ax'''  +  Bx'y'  4-  Cy'^  ^-2Ah 
+  Bk 


x'  ~h  Bh 

y'  +  Ah' 

+  2Ck 

-\-Bhk 

+  E 

+  Ck^ 

-\-Dh 

■\-Ek 

+  F 

0. 


I 


TRANSFORMATION  OF  COORDINATES  171 

Setting  equal  to  zero  the  coefficients  of  x'  and  y\  we  obtain 

(6)  2Ah-{-  Bk-{-  D  =  (), 

(7)  Bh  +  2Ck^E  =  0. 

These  equations  can  be  solved  for  h  and  k  unless  (Theorem 

IV,  p.  90) 

2A_B 

B    ~2C 
or  B'^-^AC  =  0. 

If  the  values  obtained  be  substituted  in  (5),  the  resulting  equa- 
tion will  not  contain  the  terms  of  the  first  degree.  q.e.d. 

Corollary  I.  If  an  equation  of  the  second  degree  he  transformed 
by  translating  the  axes,  the  coefficients  of  the  terms  of  the  second 
degree  are  unchanged  unless  the  new  equation  be  multiplied  or 
divided  by  some  constant. 

For  these  coefficients  in  (5)  are  the  same  as  in  the  given  equation. 

Corollary  II.  When  A  is  not  zero  the  locus  of  an  equation  of  the 
second  degree  has  a  center  of  symmetry. 

For  if  the  terms  of  the  first  degree  be  removed  the  locus  will  be  symmetrical 
with  respect  to  the  new  origin  (Theorem  V,  p.  73) . 

If  A=  ^2_4  j[(7—  0,  equations  (6)  and  (7)  may  still  be  solved  for  h  and  k 

2A       B       D 
if  (Theorem  IV,  p.  90)  -:—  =  — ;  =  —  >  when  the  new  origin  {h,  k)  may  be  any 

point  on  the  line  2Ax-\-  By  -\-  D  =  0.  In  this  case  every  point  on  that  line  will 
be  a  center  of  symmetry. 

For  example,  consider  x2  +  4a;^/  +  4?/2-|-4a;  +  8?/  +  3  =  0.  For  this  equation 
equations  (6)  and  (7)  become 

In  these  equations  the  coefficients  are  all  proportional  and  there  is  an  infinite 
number  of  solutions.  One  solution  is  A  =  —  2,  A:  =  0.  For  these  values  the  given 
equation  reduces  to 

cc2  +  4  xy  +  4  ?/2  —  1  =  0, 
or  (x  +  2  y  +  1)  (x  +  2  2/  —  1)  =  0. 

The  locus  consists  of  two  parallel  lines  and  evidently  is  symmetrical  with 
respect  to  any  point  on  the  line  midway  between  those  lines. 


172  ANALYTIC   GEOMETRY 

MISCELLANEOUS  PROBLEMS  '^ 

1.  Simplify  and  plot. 

(a)  y^-6y  +  6  =  0.  (e)  x^  +  4xy  +  y"^  =  S. 

(b)  x2  +  2xy  +  2/2  -  6x  -  6i/  +  5  =  0.     (f)  x^  -  9 y^  -  2 x  -  S6y  +  4  =  0. 
X{c)  ?/2  +  6x- 102/4-2  =  0.  (g)  25  2/2 -16x2 +  502/ -119  =  0. 

(d)  x2  +  42/2  _  8x  -  \Qy  =  0.  (h)  x2  +  2x2/  +  ?/2  _  Sx  =  0. 

2.  Find  the  point  to  which  the  origin  must  be  moved  to  remove  the  terms 
of  the  first  degree  from  an  equation  of  the  second  degree  (Theorem  VII). 

3.  To  what  point  (h,  k)  must  we  translate  the  axes  to  transform 

(1  -  e2) x2  +  2/2  -  2px  +  p2  =  0  into  (1  -  e2) x2  +  2/2  -  2  e'^px -  e'^p^  =  0  ? 

4.  Simplify  the  second  equation  in  problem  3. 

6.  Derive  from  a  figure  the  equations  for  rotating  the  axes  through  +  — 
■jt  2 

and 1  and  verify  by  substitution  in  (II),  p.  162. 

2 

6.  Prove  that  every  equation  of  the  first  degree  may  be  transformed  into 
2/'  z=  0  by  moving  the  axes.     In  how  many  ways  is  this  possible  ? 

7.  The  equation  for  rotating  the  polar  axis  through  an  angle  0  is 
6  =  6'  -V  <t>. 

8.  The  equations  of  transformation  from  rectangular  to  polar  coordi- 
nates, when  the  pole  is  the  point  (^,  k)  and  the  polar  axis  makes  an  angle  of 
0  with  the  X-axis,  are 

X  =  h  +  p  cos  {6  +  0), 
y  =  k  +  psm{6  -\-  (p). 

9.  The  equations  of  transformation  from  rectangular  coordinates  to 
oblique  coordinates  are 

X  =  x'  +  2/^  cos  w, 
y  =  y'  sin  w, 
if  the  X-axes  coincide  and  the  angle  between  OX'  and  OY'  is  w. 

10.  The  equations  of  transformation  from  one  set  of  oblique  axes  to  any 
other  set  with  the  same  origin  are 

,  sin  (w  —  0)         ,  sin  (w  —  ^) 

X  =  x' — ^^ ^  +  y ^^ ^» 

sm  w  sin  w 

,  sin  0  .     ,  sin  \Li 

sm  oj  sm  (a 

where  w  is  the  angle  between  OX  and  OF,  0  is  the  angle  from  OX  to  OX', 

and  \p  is  the  angle  from  OX  to  OY'. 


CHAPTER  yill 
CONIC  SECTIONS  AND  EQUATIONS  OF  THE  SECOND  DEGREE 

72.  Equation  in  polar  coordinates.  The  locus  of  a  point  P  is 
called  a  conic  section*  if  the  ratio  of  its  distances  from  a  fixed 
point  F  and  a  fixed  line  DD  is  constant.  F  is  called  the  focus, 
DD  the  directrix,  and  the  constant  ratio  the  eccentricity.  The 
line  through  the  focus  perpendicular  to  the  directrix  is  called 
the  principal  axis. 

Theorem  I.  If  the  pole  is  the  focus  and  the  polar  axis  the  princi- 
pal axis  of  a  conic  section^  then  the  polar  equation  of  the  conic  is  •* 


(I) 


P  = 


ep 


1  —  e  cos  ^ 

where  e  is  the  eccentricity  and  p  is  the  distance  from  the  directrix 
to  the  focus. 


P(P'0) 


Froof.    Let  P  be  any  point  on  the  conic.     Then,  by  definition, 

FP 
EP  =  '- 

From  the  figure,  FP  =  p 

and  EP  =  HM  =  p  -{-  p  cos  0. 

Substituting  these  values  of  FP  and     ^^ 
EP,  we  have 


or,  solving  for  p, 


p  +  p  cos  0 

ep 


e; 


p  — 


cos  0 


Q.E.D. 


*  Because  these  curves  may  be  regarded  as  the  intersections  of  a  cone  of  revolution 
with  a  plane. 

173 


174 


ANALYTIC  GEOMETRY 


From  (I)  we  see  that 

1.  A  conic  is  symmetrical  with  respect  to  the  principal  axis. 


For  substituting 
cos  (—  6)  =  cos  d. 


6  for  6  changes  only  the  form  of  the  equation,  since 


2.  In  plotting,  no  values  of  6  need  be  excluded. 
The  other  properties  to  be  discussed  (p.  151)  show  that  three 
cases  must  be  considered  according  as  e  =  1. 
The  parabola  e  =  1.    When  e  =  1,  (I)  becomes 


P 

p  = J 

^      1  -  cos  ^ 


and  the  locus  is  called  a  parabola. 
1.  For  ^  =  0  p  =  00,    and    for   0 


TT    p  = 


V 


The  parabola 


therefore  crosses  the  principal  axis  but  once  at  the  point  0, 

called  the  vertex,  which  is  ^  to  the  left  of  the  focus  F,  or  mid- 
way between  F  and  BD. 

2.  p  becomes  infinite  when  the  denominator,  1  —  cos  0,  vanishes. 
If  1  —  cos  ^  =  0,  then  cos  0  =  1',  and  hence.  ^  =  0  is  the  only 
value  less  than  2  tc  for  which  p  is  infinite. 

77" 

/^.^^  3.  When  0  increases  from  0  to  — > 

Z 

then        cos  6  decreases  from  1  to  0, 
1  —  cos  6  increases  from  0  to  1, 
p  decreases  from  oo  to  j), 
and  the  point  P  (p,  6)  describes  the  parabola 
from  infinity  to  B. 

TT 

When  6  increases  from  —  to  tt, 

then  cos  0  decreases  from  0  to  —  1, 

1  —  cos  Q  increases  from  1  to  2, 

IP 
p  decreases  from  />  to  ^) 

and   the  point  P  {p,  6)   describes  the  parabola  from  B  to  the 
vertex  0, 


CONIC  SECTIONS 


175 


On  account  of  the  symmetry  with  respect  to  the  axis,  when 
6  increases  from  tt  to  —^i  P(p,_0)  describes  the  parabola  from  0 
to  B'-y  and  when  6  increases  from  —^  to  2  tt,  from  B'  to  infinity. 

When  e  <  1  the  conic  is  called  an  ellipse,  and  when  e  >  1, 
an  hyperbola.  The  points  of  similarity  and  difference  in  these 
curves  are  brought  out  by  considering  them  simultaneously. 


TJie  ellipse,  e  <  1 . 

ep  e 


The  hyperbola,  e  >  1. 


1.  For  ^  =  0  ^ 


1.  For  6^  =  0  p  = 


ep 


P- 


1  —  el  —  e  1  —  el  —  e 

As  e  <  1,  the  denominator,  and  hence  As  e  >  1,  the  denominator,  and  hence 

p,  is  positive,  so  that  we  obtain  a  point  p,  is  negative,  so  that  we  obtain  a 

A  on  the  ellipse  to  the  right  of  F.  point  A  on  the  hyperbola  to  the  left 

e    •>  of  F. 

As =  1  when  e<l,  according  as  e 

1-e  <  g 

^  - ,  then  FA  may  be  greater,  equal  to,  or  "^^  ^  >  ^  (numerically)  when  e  >  1^ 

less  than  FH.  ^^^^  P  >P  >  ^^  ^  ^^^  ^^  *^®  ^®^*  ^^  ^' 


T^      n  ep  e 

For  ^  =  ;r  p=:;-f— =  :^— p.    pis 


^     ^  ep  e 

YoT  d  =  7t  p  = —^—  = p.    pis 

1+e      1+e 


1+e      1+e 

positive,  and  hence  we  obtain  a  point      positive,  and  hence  we  obtain  a  sec- 
A'  to  the  left  of  F.  ond  point  A'  to  the  left  of  F. 


As   <1,   then   p<p;    so  A^  lies 

1  +  e 

between  H  and  F. 

A  and  A'  are  called  the  vertices  of 
the  ellipse. 


As  <1,   then  p<p;    so   A'   lies 

1  +  e 
between  H  and  F. 

A  and  A^  are  called  the  vertices  of 
the  hyperbola. 


176 


ANALYTIC  GEOMETRY 


The  ellipse^  e  <  1. 

2.  p  becomes  infinite  if 

1  —  e  cos  ^  =  0, 

or  cos  6  =  -• 

e 

As  e<l,  then  ->1;   and  hence 
e 
there  are  no  values  of  6  for  which 
p  becomes  infinite. 

3.  When 


then 


.6  increases  from  0  to 


cos  6  decreases  from  1  to  0, 


1  —  e  cos  6  increases  from  1  —  e  to  1 ; 

hence      p  decreases  from to  ep, 

1  —  e 

and  P  {p,  6)  describes  the  ellipse  from 

^toC. 


The  hyperbola^  e  >  1. 
2.  p  becomes  infinite  if 
1  —  e  cos  ^  =  0, 
or  cos  6  =  -• 


As  e>l,  then  -<1;  and  hence 
e 
there  are  two  values  of  6  for  which 
p  becomes  infinite. 

3.  When 
6  increases  from  0  to  cos-i  (  -  )» 

then  cos  6  decreases  from  1  to  - , 

e 

1  —  e  cos  6  increases  from  1  —  e  to  0 ; 

exi 

hence     p  decreases  from to  —  oo, 

1  — e 

and  P  (p,  6)  describes  the  lower  half 

of  the  left-hand  branch  from  A  to 

infinity. 

When 


e  increases  from  cos-i 


(i)-f- 


When  6  increases  from  —  to  7t, 
2 
then  cos  6  decreases  from  0  to  —  1, 
1  —  e  cos  d  increases  from  1  to  1  +  e ; 

hence       p  decreases  from  ep  to  — — , 

1  +  e 
and  P  (p,  d)  describes  the  ellipse  from 
C  to  A\ 

The  rest  of  the  ellipse,  A'C'A, 
may  be  obtained  from  the  symmetry 
with  respect  to  the  principal  axis. 

The  ellipse  is  a  closed  curve. 


then      cos  d  decreases  from  -  to  0, 
e 

1  —  e  cos  ^  increases  from  0  to  1 ; 

hence  p  decreases  from  oo  to  ep, 

and  P  (p,  6)  describes  the  upper  part  of 

the  right-hand  branch  from  infinity 

toC. 

When  6  increases  from  —  to  vt, 
2 
then  cos  6  decreases  from  0  to  —  1, 
1  —  e  cos  ^  increases  from  1  to  1  +  e ; 


hence       p  decreases  from  ep  to 


ep 

l  +  ~e 

and  P  (p,  6)  describes  the  hyperbola 
from  C  to  A\ 

The  rest  of  the  hyperbola,  A'C 
to  infinity  and  infinity  to  A,  may  be 
obtained  from  the  symmetry  with 
respect  to  the  principal  axis. 

The  hyperbola  has  two  infinite 
branches. 


CONIC  SECTIONS  177 

PROBLEMS 

1.  Plot  and  discuss  tlie  following  conies.     Find  e  and  p,  and  draw  the 
focus  and  directrix  of  each. 

^  3 

(e)  P 


(a)  p  = 

T- 

-  cos  5 

(b)  p  = 

2 

1- 

-  i  cos  (? 

(C)    p  = 

8 

1- 

-  2  cos  ^ 

IA\    ^  — 

5 

(g)  p  = 
(h)  p  = 


3  -  cos  0 
6 

2  -  3cos^' 

2 
2  -  cos^" 
12 


2  -  2  cos  ^  3  -  4  cos  ^ 

2.  Transform  the  equations  in  problem  1  into  rectangular  coordinates, 
simplify  by  the  Rule  on  p.  165,  and  discuss  the  resulting  equations.  Find 
the  coordinates  of  the  focus  and  the  equation  of  the  directrix  in  the  new 
variables.  Plot  the  locus  of  each  equation,  its  focus,  and  directrix  on  the 
new  axes. 

Ans.   (a)  ?/2=:4x,  (1,0),  x  =  -  1.  ." 

x2      y2      ,     /      4    ^\  16 


^    '     6_4  IJL  '     V         3  / 


3       /  3 


^    '    6_4  6_4  '     V  3  J  3 

(d)  2/2  =  5x,  (1,0),  x  =  -%. 

x2      ?/2      ^     /      3     -\  27 


64  8 

/2      .     /18     ^\  8 

5' 


w5-f-.  (-?■»)• —I 

x2         ?/2       ,     /48    ^\  2"; 

^    '   1296        144  '    V  7  /  7 


3.  Transform  (I)  into  rectangular  coordinates,  simplify,  and  find  the  coor- 
dinates of  the  focus  and  the  equation  of  the  directrix  in  the  new  rectangular 
coordinates  if  (a)  e  =  1,  (b)  e^l. 

Am.  (a)  2/2  =  2px,  (|,  o),  a:=-| 


(b) 


e2p  

(1  -  e2)2      1  _  e2 


+  _^.l,  (-^,  0),x  = 
^    e2p2         'V      1  _  e2       / 


178  ANALYTIC  GEOMETRY 

4.  Derive  the  equation  of  a  conic  section  when  (a)  the  focus  lies  to  the 
left  of  the  directrix ;  (b)  the  polar  axis  is  parallel  to  the  directrix. 

Ans.  (a)  p  = ^^ ;   (b)  p  -        ^^ 


1  +  e  cos  6  1  —  e  sin  0 

5.  Plot  and  discuss  the  following  conies.     Find  e  and  p,  and  draw  the 
directrix  of  each. 

7 


(^)  P  =  ^;-, ^-  (^)  P 


1  +  cos  e  3  +  10  cos  ^ 

5 


{h)p  =  - — ..  {d)p  = 


1  -  sin  ^  3  -  sin  e 

73.  Transformation  to  rectangular  coordinates. 

Theorem  II.    If  the  origin  is  the  focus  and  the  X-axis  the  princi- 
pal axis  of  a  conic  section,  then  its  equation  is 
(II)  (1  -  e^)  0^2  _^  ^2  _  2  e'^px  -  eY  =  0, 

where  e  is  the  eccentricity  and  xz=z—p  is  the  equation  of  the 
directrix. 

Proof    Clearing  fractions  in  (I),  p.  173,  we  obtain 

P  —  ep  cos  6  =  ep. 
Set  p  =  ±  Vx^  +  y^  and  p  cos  ^  =  a;  (p.  155).     This  gives 
±  V a;^  -{-  y^  —  ex  =  ep, 
or  ±  Vx^  +  y^  =  ex  -\-  ep. 

Squaring  and  collecting  like  pov^ers  of  x  and  y,  we  have  the 
required  equation.  Since  the  directrix  DD  (Fig.,  p.  173)  lies  p 
units  to  the  left  of  F  its  equation  is  ic  ——p.  q.e.d. 

74.  Simplification  and  discussion  of  the  equation  in  rectangu- 
lar coordinates.     The  parabola,  e  =  1. 

When  6  =  1,  (II)  becomes 

y^  —  2px  —p^  =  0. 
Applying  the  Eule  on  p.  165,  we  substitute 

(1)  x  =  x'  -\-  h,  y  =  y'  +  k, 
obtaining 

(2)  y'^  -  2px'  +  2ky'-{-k:'-  2ph  -p^  =  0. 


CONIC  SECTIONS 


179 


Set  the  coefficient  of  y^  and  the  constant  term  equal  to  zero 
and  solve  for  h  and  h.     This  gives 


(3) 


7,=-|,    7,  =  0. 


Substituting  these  values  in  (2)  and  dropping  primes,  the  equa- 
tion of  the  parabola  becomes  y^  =  2jjx. 

From  (3)  we  see  that  the  origin  has  been 
removed  from  F  to  O,  the  vertex  of  the    ^ 
parabola.     It  is  easily  seen  that  the  new 

coordinates  of  the  focus  are  [  ^ ,  0  1 ,  and    x 

the    new   equation    of    the    directrix    is 
X  =  —  ^'    Hence 

Li 

Theorem  III.    Jf  the  origin  is  the  vertex  and  the  X-axis  the  axis 
of  a  parabola,  then  its  equation  is 

(III)  y^  =  2poe. 

The  focus  is  the  point  (  — ,  0  ),  and  the  equation  of  the  directrix 

V  \        I 

IS  X  =  —  —• 


A  general  discussion  of  (III)  gives  us  the  following  properties  of  the 
parabola  in  addition  to  those  already  obtained 
(p.  174). 

1.  It  passes  through  the  origin  but  does  not  cut 
the  axes  elsewhere. 

2.  Values  of  x  having  the  sign  opposite  to  that 
X    of  p  are  to  be  excluded  (Rule,  p.  73).     Hence  the 

curve  lies  to  the  right  of  YY'  when  p  is  positive  and 
to  the  left  when  p  is  negative. 

3.  No  values  of  y  are  to  be  excluded ;  hence  the 
curve  extends  indefinitely  up  and  down. 

Theorem  IV.    If  the  origin  is  the  vertex  and  the  Y-axis  the  axis 
of  a  parabola,  then  its  equation  is 


(IV) 


a?2  =  2py. 


180 


ANALYTIC  GEOMETRY 


V 

7 

\        / 

\ 

^ 

V , 

X' 

0 

V-^    X 

D 

Y' 

D 

The  focus  is  the  point  (  0,  "^  I  j  and  the  equation  of  the  directrix 
.  V 

Proof  Transform  (HI)  by  rotating  the 

77" 

axes  through  —  —  •  Equations  (II),  p.  162, 

give  us  for  B  =~  — 

x  =  y\ 
y=-x'. 
Substituting  in  (III)  and  dropping  primes,  we  obtain  x^  =  2py. 

Q.E.D. 

After  rotating  the  axes  the  whole  figure  is 

turned  through  —  in  the  positive  direction. 

The  parabola  lies  above  or  below  the  X-axis 
according  as  p  is  positive  or  negative. 

Equations  (III)  and  (IV)  are  called 
the  typical  forms  of  the  equation  of  the 
parabola. 

Equations  of  the  forms 

Ax^  +  Ey  =  0  and   Cy^  +  Dx  =  0, 

where  A,  E,  C,  and  D  are  different  from  zero,  may,  by  transpo- 
sition and  division,  be  written  in  one 

^  of  the  typical  forms  (III)  or  (IV), 
so  that  in  each  case  the  locus  is  a 

^  parabola. 

Ex.  1.    Plot  the  locus  of  x^  +  4  ?/  =  0  and 
find  the  focus  and  directrix. 

Solution.    The  given  equation  may  be 
written 

x2  =  -  4  y. 


Y' 

'y-f 

D 

^        P 

X' 

y^o 

^\      X 

ya 

^ 

D 

y^i 

0 

X' 

/^ 

"^ 

N 

/ 

^(C 

,-i; 

\ 

/ 

\ 

1 

\ 

i 

i 

1 

\ 

Y' 

Comparing  with  (IV),  the  locus  is  seen  to  be  a  parabola  for  which  p  =  —2. 
Its  focus  is  therefore  the  point  (0,  —  1)  and  its  directrix  the  line  y  =  1. 

Ex.  2.    Find  the  equation  of  the  parabola  whose  vertex  is  the  point  0' 
(3,  —  2)  and  whose  directrix  is  parallel  to  the  F-axis,  if  p  =  3. 


CONIC  SECTIONS 


181 


Solution.    Referred  to  O'X'  and  O'Y'  as  axes,  the  equation  of  the  parabola 
is  (Theorem  III) 

(4)  y'^  =  ex\ 
The  equation  for  translating  the  axes  from 

0  to  (7  are  (Theorem  I,  p.  IGO) 

X  =  x'  +  3,  y  =  y'  -2, 
whence 

(5)  x'  =  x-^,y'  =  y  +  2. 
Substituting  in  (4),  we  obtain  as  the  re- 
quired equation 

(2/ +  2)2  =  6(^-3), 
or  2/2_6a.^4y  ^22  =  0. 

Referred  to  O'X'  and  0'Y\  the  coordinates 
of  T  are  (Theorem  III)  (|,  0)  and  the  equa- 
tion of  DB  is  x'  =  —  |.     By  (5)  we  see  that, 
referred  to  OX  and  OF,  the  coordinates  of  F  are  (|,  -  2)  and  the  equation 
of  DD  is  X  =  ir 


nf 

rV 

> 

^ 

y 

/ 

/ 

0 

5: 

F 

.O' 

b,- 

2) 

.Y 

. 

s. 

V 

s 

^ 

Z.I 

PROBLEMS 

1 .  Plot  the  locus  of  the  following  equations.     Draw  the  focus  and  direc- 
trix in  each  case. 


(a)  y^  =  4:X. 

(b)  2/2 +  4x 


{6)  y^  -6x  =  0. 
0.  (e)  x2  +  10  y  =  0. 

(c)  x2-82/  =  0.  (f)  2/2  +  x  =  0. 

2.  If  the  directrix  is  parallel  to  the  F-axis,  find  the  equation  of  the 
parabola  for  which 

(a)  p  =  6,  if  the  vertex  is  (3,  4). 

(b)  p  =  —  4,  if  the  vertex  is  (2,  —  3). 

(c)  p  =  8,  if  the  vertex  is  (—5,  7). 

(d)  p  '=  4,  if  the  vertex  is  (/i,  k). 

3.  The  chord  through  the  focus  perpendicular  to  the  axis  is  called  the  latus 
rectum.      Find  the  length  of  the  latus  rectum  of  y^  =  2px.  Ans.   2  p. 

4.  What  is  the  equation  of  the  parabola  whose  axis  is  parallel  to  the  axis 
of  y  and  whose  vertex  is  the  point  (or,  )3)  ?       Ans.   (x  —  a)^  =  2p{y  —  ^). 

5 .  Transform  to  polar  coordinates  and  discuss  the  resulting  equations 
(a)  2/2  =  2px,  (b)  x2  =  2py. 

6.  Prove  that  the  abscissas  of  two  points-on  the  parabola  (III)  are  propor- 
tional to  the  squares  of  the  ordinates  of  those  points. 


Ans.  (?/ -  4)2  =  12  (X  -  3). 

Ans.  (2/  +  3)2=-8(x-2). 

Ans.  {y  -  7)2  =  16  (x  +  5). 

Ans.  {y  -kf  =  8{x-  h). 


182  ANALYTIC   GEOMETRY 

75.  Simplification  and  discussion  of  the  equation  in  rectan- 
gular coordinates.  Central  conies,  e%l.  When  e^l,  equation 
(II),  p.  178,  is 

(1  -  e2)a;2  +  2/2-2  e''2)x  -  eY  =  0. 

To  simplify  (Rule,  p.  165),  set 

(1)  x  =  x'  -{-h,  y  =  y'  +  k, 
which  gives 

(2)  (1  -  e^)x'''  +  2/''  -f  2  h  (1  -  e^) 


=  0. 


ic'  + 2  7^2/' +  (1-6^)^2 

-  2  e'^ph 

—  e^ 

Setting  the  coefficients  of  cc'  and  y^  equal  to  zero  gives 

2  hi\  -e^)-2e^p  =  0,  2k  =  0,     '^  - 
whence 

(3)  _       *  =  r^'  *  =  o- 

Substituting  in  (2)  and  dropping  primes,  we  obtain 
or 

(4)  — r^  +  -fr  =  i- 

(1  -  6^)2  1-^2 

This  is  obtained  by  transposing  the  constant  term,  dividing  by  it,  and  then 
dividing  numerator  and  denominator  of  the  first  fraction  by  1  —  e^. 

The  ellipse,  e  <  1.  The  hyperbola,  e  >  1. 

From  (3)  it  is  seen  that  h  is  posi-  From  (3)  it  is  seen  that  h  is  nega- 

tive when  e  <  1.     Hence  the  new  ori-      tive  when  e  >  1.     Hence  the  new  ori- 
gin 0  lies  to  the  right  of  the  focus  F.      gin  0  lies  to  the  left  of  the  focus  F. 

Further, >  1  numerically,  so 

'  1  -  e2 

h>p  numerically ;   and  hence  the 

new  origin  lies  to  the  left  of  the 

directrix  DD. 


CONIC  SECTIONS 


183 


The  locus  of  (4)  is  symmetrical  with  respect  to  YY'  (Theorem 
V,  p.  73).     Hence  0  is  the  middle  point  oi  AA'.     Construct  in 


D 

Y 

d' 

E 

-A 

B^ 

^t 

e' 

^?Sv 

1 

\ 

\ 

\ 

X' 

A 

F 

0 

F' 

y 

X 

D 

V. 

Y 

B^ 

y 

D' 

either  figure  F'  and  D'D'  symmetrical  respectively  to  F  and  DB 
with  respect  to  YY\     Then  F'  and  X)'i)'  are  a  new  focus  and  ^ 
directrix. 

For  let  P  and  P'  be  two  points  on  the  curve,  symmetrical  with  respect  to  YY'. 

Then  from  the  symmetry  PF  —  P'F'  and  PE  —  P'E'.    But  since,  by  definition, 

PF  P'F' 

— —  =  e,  then  =  e.    Hence  the  same  conic  is  traced  by  P',  using  F'  as  focus 

PMi  P  Mi 

and  D'D'  |is  directrix,  as  is  traced  by  P,  using  F  as  focus  and  DD  as  directrix. 

Since  the  locus  of  (4)  is  symmetrical  with  respect  to  the  origin 
(Theorem  V,  p.  73),  it  is  called  a  central  conic,  and  the  center  of 
symmetry  is  called  the  center.  Hence  a  central  conic  has  two  foci 
and  two  directrices. 

The  coordinates  of  the  focus  F  in  either  figure  are 


(-A») 


For  the  old  coordinates  of  F  were  (0,  0).    Substituting  in  (1),  the  new  coordi- 
nates are  x  =  —  h,  y'  =  —  k,  or,  from  (3),  /  —  - — ^.  0  V 


i 


The  coordinates  of  F'  are  therefore 


The  new  equation  of  the  directrix  DD  is  x 


-      _^ 


184 


ANALYTIC  GEOMETRY 


For    from    (1)    and    (3),   x  =  x'  + 


e^ 


y  =  y 


l-e2 
(Theorem  II)  and  dropping  primes,  we  obtain  x  = 

P 


Substituting  in  x  =  —p 
P 


Hence  the  equation  oi  D'D'  is  x  =  j 

We  thus  have  the 

Lemma.    The  equation  of  a  central  conic  whose  center  is  the  origin 
and  whose  principal  axis  is  the  X-axis  is 

(4)  _^!_  +  ^L  =  l. 


e^p^ 


Its  foci  are  the  points     I  ± 


e^p 


e^p^ 
0^ 


1-e' 


and  its  directrices  are  the  lines  x  =  :t 


IP 


The  ellipse,  e  <  1. 
For  convenience  set 


(5)  a 


l-e2  1 


e^p 


a2  and  6^  are  the  denominators  in  (4) 
and  c  is  the  abscissa  of  one  focus.    Since 
-.e<l,  l-e2  is  positive;   and  hence  a,  b^, 
and  c  are  positive. 


We  have  at  once 


a^-b'^  = 


e^p-^ 


e^p-^ 


(1 


e4p2 


and 


(1  -  e2)2 


e-^p-' 


e^p 


(1  -  e2)2      1  _  e2      1 


Hence  the  directrices  (Lemma)  are 

the  lines  x  =  ±—- 
c 

By  substitution  from  (5)  in  (4)  we 

obtain 

a2  "*■  62       " 


The  hyperbola,  e  >  1. 
Eor  convenience  set 


(6) 
a  = 


ep 
l-e2' 


62: 


e^p 


a^  and  -  b^  are  the  denominators  in  (4) 
and  c  is  the  abscissa  of  one  focus.  Since 
e  >  1, 1-  e2  is  negative;  and  hence  a,  b^,  and 
c  are  positive. 


We  have  at  once 


a2  +  62 

e2p2 

(1  -  e2)2 

e4p2 

(1  -  e2)2 

e2p2 

.         e2p 

2ri2 


a^p 


l-e2 
c2 


and 

a2__ 
c  ~  (1  -  e2)2 


l-e2 


Hence  the  directrices  (Lemma)  are 

the  lines  x  =  ±  —• 
c 

By  substitution  from  (6)  in  (4)  we 

obtain 

X2_y2^_^ 

a2      62 


CONIC  SECTIONS 


185 


The  ellipse^  e  <  1. 
The  intercepts  are  x  =  ±a  and 
y  =  ±h.  AA'  =  2  a  is  called  the 
major  axis  and  BB'  =  26  the  minor 
axis.  Since  a^  —  h^  =  c^  is  positive, 
then  a>b,  and  the  major  axis  is 
greater  than  the  minor  axis. 


^1 

^ 

D 

/ 

/" 

h 

B 

D 

1 

V 

<—a-\ — * 

X' 

A 

F 

0 

^.-C—^F'JA 

X 

D 

\ 

Y 

f____«^____. 

d' 

Hence  we  may  restate  the  Lemma 
as  follows. 

Theorem  V.  The  equation  of  an 
ellipse  whose  center  is  the  origin  and 
whose  foci  are  on  the  X-axis  is 


(V) 


a' 


where  2  a  is  the  major  axis  and  2  b  the 
minor  axis.  If  c'^  =  a^  —  b^,  then  the 
foci  are  (±  c,  0)  and  the  directrices 

are  x  =  ±  —- 
c 

Equations  (5)  also  enable  us  to 

express  e  and  p,  the  constants  of  (I), 

p.  173,  in  terms  of  a,  6,  and  c,  the 

constants  of  (V).     For 

c  _    e^p  ev     _ 

a~l-e2"l-e2~ 


(7) 

and 

(9) 


62 

c 


l-e2 


e^p 


P 


The  hyperbola,  e  >  1. 
The  intercepts  are  x  =  ±  a,  but  the 
hyperbola  does  not  cut  the  F-axis. 
AA'  =  2  a  is  called  the  transverse 
axis  and  BB'  =  26  the  conjugate 
axis. 


Hence  we  may  restate  the  Lemma 
as  follows. 

Theorem  VI.  The  equation  of  an 
hyperbola  whose  center  is  the  origiru 
and  whose  foci  are  on  the  X-axis  is 


x^ 


r 


a^      o2 

where  2  a  is  the  transverse  axis  and  2  b 
the  conjugate  axis.  If  c^  =  a^-\-  b^, 
then  the  foci  are  (±  c,  0)  and  the 

directrices  are  x  =  i  —  • 
c 

Equations  (6)  also  enable  us  to 

express  e  and  p,  the  constants  of  (I), 

p.  173,  in  terms  of  a,  6,  and  c,  the 

constants  of  (VI).     For 

(8)  1^        '"P     .  ^     -- 


and 


l-e2 
e2p2 


(10)    —  = 

^     ^    c  l-e2 


l-e2 

e^p 
l-e2 


=  1>. 


186 


ANALYTIC  GEOMETRY 


The  ellipse,  e  <  1. 

In  the  figure  OB  =  b,  OF"  =  c, 
and  since  c^  =  a'^  —  6^,  then  BF'  =  a. 
Hence  to  draw  the  foci,  with  JB  as  a 
center  and  radius  OA,  describe  arcs 
cutting  XX'  at  F  and  F'.  Then  F 
and  F'  are  the  foci. 

It  a  =  b,  then  (V)  becomes 

x2  +  2/2  =  a2, 

whose  locus  is  a  circle. 


Transform    (V)   by    rotating   the 

axes  through  an  angle  of (Theo- 

2 

rem  II,  p.  162).     We  obtain 

Theorem  VII.     The  equation  of  an 
ellipse  whose  center  is  the  origin  and 
whose  foci  are  on  the  Y-axis  is 
«2_^^2 

r2 


The  hyperbola,  e>l. 

In  the  figure  OB  =  6,  OA'  =  a ; 
and  since  c2  =  a2  +  62^  then  BA'  =  c. 
Hence  to  draw  the  foci,  with  0  as  a 
center  and  radius  BA',  describe  arcs 
cutting  XX'  at  F  and  F'.  Then  F 
and  F'  are  the  foci. 

If  a  =  6,  then  (VI)  becomes 
x2-y2  =  a2, 

whose  locus  is  called  an  equilateral 
hyperbola. 

Transform  (VI)  by  rotating  the 

axes  through  an  angle  of (Theo- 

rem  II,  p.  162).     We  obtain 

Theorem  Vni.  The  equation  of  an 
hyperbola  whose  center  is  the  origin 
and  whose  foci  are  on  the  Y-axis  is 


(VII) 


(VIII)         ^  —  JL-IL 


=  1, 


Y 

\ 

B' 

A 

B'\ 

i 

f          / 
/    a/ 

y 

/ 

F' 

c 

A2 

X'  n      10 

F 

\B'      £ 

D 

D 

Y 

where  2  a  is  the  major  axis  and  2  bis 

the  minor  axis.     If  d^  =  a^  —  62,  the 

foci  are  (0,  ±  c)  and  the  directrices 

a2 
are  the  lines  y  =±  —• 


where  2a  is  the  transverse  axis  and  2h 
is  the  conjugate  axis.  If  c'^  =  a^  +  b^, 
the  foci  are  (0,  ±  c)  and  the  directrices 

are  the  lines  y  =  ±^. 


CONIC  SECTIONS 


187 


The  ellipse^  e  <  1. 

The  essential  difference  between 
(V)  and  (VII)  is  that  in  (V)  the  de- 
nominator of  x2  is  larger  than  that 
of  2/^,  while  in  (VII)  the  denominator 
of  y^  is  the  larger.  (V)  and  (VII) 
are  called  the  typical  forms  of  the 
equation  of  an  ellipse. 


The  hyperbola^  e  >  1. 

The  essential  difference  between 
(VI)  and  (VIII)  is  that  the  coeffi- 
cient of  ?/2  is  negative  in  (VI),  while 
in  (VIII)  the  coefficient  of  x^  is  nega- 
tive. (VI)  and  (VIII)  are  called  the 
typical  forms  of  the  equation  of  an 
hyperbola. 


An  equation  of  the  form 


where  A,  C,  and  F  are  all  different  from  zero,  may  always  be 
written  in  the  form 


(11) 


a       /3 


By  transposing  the  constant  term  and  then  dividing  by  it,  and  dividing 
numerator  and  denominator  of  the  resulting  fractions  by  A  and  C  respectively. 

The  locus  of  this  equation  will  be 

1.  An  ellipse  if  a  and  (3  are  both  positive,  a^  will  be  equal  to 
the  larger  denominator  and  b^  to  the  smaller. 

2.  An  hyperbola  if  a  and  ^  have  opposite  signs,  a^  will  be 
equal  to  the  positive  denominator  and  b^  to  the  negative  denomi- 
nator. 

3.  If  a  and  p  are  both  negative,  (11)  will  have  no  locus. 

Ex.  1.    Find  the  axes,   foci,  directrices,  and 
eccentricity  of  the  ellipse  4  x^  +  y2  =  16. 

Solution.    Dividing  by  16,  we  obtain 

-  +  ^  =  1 
4       16 

The  second  denominator  is  the  larger.     By 
comparison  with  (VII), 

62  =  4,  a2  =  16,  c2  =  16  -  4  =  12. 

Hence  6  =  2,        a  =  4,       c  =  Vl2. 

The  positive  sign  only  is  used  when  we  extract  the 
square  root,  becaus.e  a,  6,'and  c  are  essentially  positive. 


r> 

, 

JJ 

D 

__/■ 

^ 

> 

V 

/ 

\ 

/ 

\ 

7? 

B' 

0 

X 

\ 

/ 

\ 

/ 

\ 

"'■i 

/ 

■a 

r\ 

T\ 

T' 

/s^       ^ 


y 


188  ANALYTIC  GEOMETRY 

Hence  the  major  axis^^^'  =  8,  the  minor  axis  BB'  =  4,  the  foci  F  and  F' 

are  the  points  (0,  ±  Vl2),  and  the  equations  of  the  directrices  DD  and  D'ly 

a2  16        ^4 

are  y  =  ±  —  =  ±  —=  =  ±  -  Vl2. 
c  Vl2  3 

Vl2  4         1 

From  (7)  and  (9),  e  =  — —  and  p  =  -— =:  =  -  V12. 
4  V12      3 


PROBLEMS 

1.  Plot  the  loci,  directrices,  and  foci  of  the  following  equations  and  find 
e  and  p. 

(a)  x2  +  9?/2  =  81.  (e)  dy^  -  4x2  =  36. 

(b)  9  x2  _  16  y2  =  144.  (f )  x2  -  2/2  =  25. 

(c)  16  x2  +  2/2  =  25.  (g)  4  x2  +  7  2/2  =  13. 

(d)  4x2  +  92/2  =  36.  (h)  5x2  _  32/2  =  14. 

2.  Find  the  equation  of  the  ellipse  whose  center  is  the  origin  and  whose 
foci  are  on  the  JT-axis  if 

(a)  a  =  5,  6  =  3.  Ans.  9  x2  +  25  2/2  =  225. 

(b)a'=6,  e=:i.  Ans.  32x2  +  362/2  =  1152. 

(c)  6  =  4,  c  =  3.  Ans.  16  x2  +  25  2/2  =  400. 

(d)  c  =  8,  e  =  f.  Ans.  5x2  +  9 2/2  =  720, 

3.  Find  the  equation  of  the  hyperbola  whose  center  is  the  origin  and 
whose  foci  are  on  the  X-axis  if 

(a)  a  =  3,  h  =  5.  Ans.  2bx^-9y^  =  225. 

(b)  a  =  4,  c  =  5.  Ans.  9x^-16y^  =  144. 

(c)  e  =  f,  a  =  5.  Ans.  5x2-42/2  =  125. 

(d)  c  =  8,  e  =  4.  Ans.  15x2-2/2  =  60. 

4.  Show  that  the  latus  rectum  (chord  through  the  focus  perpendicular  to 

2  62 

the  principal  axis)  of  the  ellipse  and  hyperbola  is 

a 

5.  What  is  the  eccentricity  of  an  equilateral  hyperbola  ?        A71S.    V2. 

6.  Transform  (V)  and  (VI)  to  polar  coordinates  and  discuss  the  resulting 
equations. 

7.  Where  are  the  foci  and  directrices  of  the  circle  ? 

8.  What  are  the  equations  of  the  ellipse  and  hyperbola  whose  centers 
are  the  point  (a,  /3)  and  whose  principal  axes  are  parallel  to  the  X-axis  ? 

Ans   ('^-^)^  ,  (y-^)^^i.  i^-o^)^    (y-/3)2^ 

a2  62  '        a2  62  * 


CONIC  SECTIONS  189 

76.  Conjugate  hyperbolas  and  asymptotes.  Two  hyperbolas 
are  called  conjugate  hyperbolas  if  the  transverse  and  conjugate 
axes  of  one  are  respectively  the  conjugate  and  transverse  axes  of 
the  other.  They  will  have  the  same  center  and  their  principal 
axes  (p.  173)  will  be  perpendicular. 

If  the  equation  of  an  hyperbola  is  given  in  typical  form,  then 
the  equatlofi  of  the  conjugate  hyperbola  is  found  by  changing  the 
signs  of  the  coefficients  of  x^  and  y'^  in  the  given  equation. 

For  if  one  equation  be  written  in  the  form  (VI)  and  the  other  in  the  form  (VIII), 
then  the  positive  denominator  of  either  is  numerically  the  same  as  the  negative 
denominator  of  the  other.  Hence  the  transverse  axis  of  either  is  the  conjugate 
axis  of  the  other. 

Thus  the  loci  of  the  equations 

(1)  16x2-2/2^16  and  -16x2  +  2/2=16 

are  conjugate  hyperbolas.    They  may  be  written 

«2      2/2  a;2      2/2 

___^la„d--  +  -  =  l. 

The  foci  of  the  first  are  on  the  X-axis,  those  of  the  second  on  the  T-axis.  The 
transverse  axis  of  the  first  and  the  conjugate  axis  of  the  second  are  equal  to  2, 
while  the  conjugate  axis  of  the  first  and  the  transverse  axis  of  the  second  are 
equal  to  8. 

The  foci  of  two  conjugate  hyperbolas  are  equally  distant  from 
the  origin. 

For  c2  (Theorems  VI  and  VIII)  equals  the  sum  of  the  squares  of  the  semi- 
transverse  and  semi-conjugate  axes,  and  that  sum  is  the  same  for  two  conjugate 
hyperbolas. 

Thus  in  the  first  of  the  hyperbolas  above  c^  =  1  -f  16,  while  in  the  second 
c2=16  +  l. 

If  in  one  of  the  typical  forms  of  the  equation  of  an  hyperbola 
we  replace  the  constant  term  by  zero,  then  the  locus  of  the  new 
equation  is  a  pair  of  lines  (Theorem,  p.  Q>^)  which  are  called  the 
asymptotes  of  the  hyperbola. 

Thus  the  asymptotes  of  the  hyperbola 

(2)  b'^x'-  -  aSf  =  a%'' 
are  the  lines 

(3)  ^2^2  _  ^2^2  ^  0, 

or 

(4)  bx  -\-  ay  =  0  and  bx  —  ay  =  0. 


190  ANALYTIC  GEOMETRY 

Both  of  these  lines  pass  through  the  origin,  and  their  slopes  are  respectively 
,5)  -^nd^. 

An  important  property  of  the  asymptotes  is  given  by 

Theorem  IX.  The  branches  of  the  hyperbola  approach  its  asymp- 
totes as  they  recede  to  infinity. 

Proof.  Let  P^  (x-^,  y\)  be  a  point  on  either  branch  of  (2)  near 
the  first  of  the  asymptotes  (4).  The  distance  from  this  line  to 
Pi  (Fig.,  p.  191)  is  (Eule,  p.  106) 

(6)  d=  ^-^±^y^. 

Since  Pi  lies  on  (2),    bW  -  «Vi'  =  «^'^'- 

Factoring,  bx^  +  a?/i  = ■_ —  "^ 

oxi  —  ayi 

a%^ 

Substituting  in  (6),     d  = .  - 

4-  VZ>2  +  a^  (bxi  —  ayi) 

As  Pi  recedes  to  infinity,  a^i  and  ?/i  become  infinite  and  d 
approaches  zero. 

For  bxi  and  ayi  cannot  cancel,  since  Xi  and  pi  have  opposite  signs  in  the  second 
and  fourth  quadrants. 

Hence  the  curve  approaches  closer  and  closer  to  its  asymptotes. 

Q.E.D. 

Two  conjugate  hyperbolas  have  the  same  asymptotes. 

For  if  we  replace  the  constant  term  in  both  equations  by  zero,  the  resulting 
equations  differ  only  in  form  and  hence  have  the  same  loci. 

Thus  the  asymptotes  of  the  conjugate  hyperbolas  (1)  are  respectively  the  loci  of 

16x2  -  2/2  =  0  and  -lGx^  +  y'^  =  0, 
which  are  the  same. 

An  hyperbola  may  be  drawn  with  fair  accuracy  by  the  fol- 
lowing 

Construction.  Lay  o&  OA  =  OA'  =  a  on  the  axis  on  which  the 
foci  lie,  and  OB  =  OB'  =  b  on  the  other  axis.  Draw  lines  through 
A,  A',  B,  B'  parallel  to  the  axes,  forming  a  rectangle.^*    Draw  the 

*  An  ellipse  may  be  drawn  with  fair  accuracy  by  inscribing  it  in  such  a  rectangle. 


CONIC  SECTIONS 


191 


diagonals  of  the  rectangle  and  the  circumscribed  circle.  Draw 
the  branches  of  the 
hyperbola  tangent  to 
the  sides  of  the  rec- 
tangle at  A  and  A' 
and  approaching  nearer 
and  nearer  to  the  di- 
agonals. The  conju- 
gate hyperbola  may 
be  drawn  tangent  to 
the  sides  of  the  rec- 
tangle at  B  and  B' 
and  approaching  the  diagonals.  The  foci  of  both  are  the  points 
in  which  the  circle  cuts  the  axes. 

The  diagonals  will  be  the  asymptotes,  because  two  of  the  vertices  of  the  rec- 
tangle ( ±  a,  ±  6)  will  lie  on  each  asymptote  (4) .    Half  the  diagonal  will  equal  c,  ^ 
the  distance  from  the  origin  to  the  foci,  because  c^=  a^  -\-  b^. 

77.  The  equilateral  hyperbola  referred  to  its  asymptotes.   The  equation 
of  the  equilateral  hyperbola  (p.  186)  is 
(1)  x2  -  2/2  =  ot2.         - 

Its  asymptotes  are  the  lines 

X  —  y  —  0  and  ic  +  y  =  0. 

These  lines  are  perpendicular  (Corollary  III,  p.  87),  and  hence  they  may 
be  used  as  coordinate  axes. 

Theorem  X.    The  equation  <^f  an  equilateral  hyperbola  referred  to  its  asymp- 
totes is 
(X)  2ocy  =  a^. 

It 
Proof.    The  axes  must  be  rotated  through 

to  coincide  with  the  asymptotes. 

Hence  we  substitute  (Theorem  II,  p.  162) 

x'  -\-y'  -x'  +  y' 


V2 


V2 
in  (1).     This  gives 


Or,  reducing  and  dropping  primes, 

1xy  =  a2. 


Q.E.IX 


192 


ANALYTIC  GEOMETRY 


78.  Focal  property  of  central  conies.  A  line  joining  a  point  on 
a  conic  to  a  focus  is  called  a  focal  radius.  Two  focal  radii,  one  to 
each  focus,  may  evidently  be  drawn  from  any  point  on  a  central 
conic. 


-  Theorem  XI,  The  sum  of  the  focal 
radii  from  any  point  on  an  ellipse  is 
equal  to  the  major  axis  2  a. 


Theorem  Xn.  The  difference  of  the 
focal  radii  from  any  point  on  an 
hyperbola  is  equal  to  the  transverse 
axis  2  a. 


T> 

Y' 

d' 

E 
II 

^ 

r"^^ 

e' 

< 

f^ 

^ 

\ 

X' 

A 

F 

0 

jr' 

1^ 

X 

D 

V 

b' 

r' 

r| 

D                  J 

E       ly 

h' 

H 

1  \r 

X'     F' JA 
/         ^ 

0 
y' 

\i  X 
D        \ 

Proof.    Let  P  be  any  point  on  the 
ellipse.     By  definition  (p.  173), 
r  =  e-  PE,  r'  =  e  .  PE\ 
Hence  r  +  r'  =  e  {PE  +  PE") 
=  e  •  HH\ 

From  (7),  p.  185,  e=  -» 


Proof.    Let  P  be  any  point  on  the 
hyperbola.     By  definition  (p.  173), 
r  =  e-  PE,  r'  =  e-  PE'. 
Hence  r'  -  r  =  e  {PE'  -  PE) 
=  e  •  HH\ 

From  (8),  p.  185,  e=  -, 


and  from  the  equations  of  the  direc- 

and from  the  equations  of  the  direc- 

trices (Theorem  V), 

trices  (Theorem  VI), 

«2 

c 

a2 
HH'  =  2  —  . 
c 

Hence  r+r'  =  -.2-  =  2a. 

«          '                 Q.E.D. 

Hence»-'-r  =  --2-  =  2a. 

«       "           Q.K.n. 

79.  Mechanical  construction  of  conies.  Theorems  XI  and  XII  afford  simple 
methods  of  drawing  ellipses  and  hyperbolas.  Place  two  tacks  in  the  drawing 
board  at  the  foci  F  and  F'  and  wind  a  string  about  them  as  indicated. 

If  the  string  be  held  fast  at  A,  and  a  pencil  be  placed  in  the  loop  FPF' 
and  be  moved  so  as  to  keep  the  string  taut,  then  PF  +  PF'  is  constant  and 
P  describes  an  ellipse.  If  the  major  axis  is  to  be  2  a,  then  the  length  of  the 
loop  FPF'  must  be  2  a. 


CONIC  SECTIONS 


193 


If  the  pencil  be  tied  to  the  string  at  P,  and  both  strings  be  pulled  in  or 
let  out  at  A  at  the  same  time,  then  PF"  —  PF  will  be  constant  and  P  will 
describe  an  hyperbola.  If  the  transverse  axis  is  to  be  2  a,  the  strings  must 
be  adjusted  at  the  start  so  that  the  difference  between  PF'  and  PF  equals  2  a. 


To  describe  a  parabola,  place  a  right  triangle  with  one  leg  EB  on  the 
directrix  DD.  Fasten  one  end  of  a  string  whose  length  is  AE  at  the  focus 
F,  and  the  other  end  to  the  triangle  at  A.  With  a  pencil  at  P  keep  the 
string  taut.  Then  PF  =  PE ;  and  as  the  triangle  is  moved  along  Di>  the 
point  P  will  describe  a  parabola. 


PROBLEMS 

1.  Find  the  equations  of  the  asymptotes  and  hyperbolas  conjugate  to  the 
following  hyperbolas,  and  plot. 

(a)  4ic2  -  2/2  =  36.  (c)  16x2  _  y2  4.  64  =  0. 

(b)  9x2  -  25y2  =  loO.  (d)  8x2  -  16y2  +  25  =  0. 

2.  Prove  Theorem  IX  for  the  asymptote  which  passes  through  the  first 
and  third  quadrants. 

3.  If  e  and  e'  are  the  eccentricities  of  two  conjugate  hyperbolas,  then 
1        1       . 

4.  The  distance  from  an  asymptote  of  an  hyperbola  to  its  foci  is  numer- 
ically equal  to  6. 

5.  The  distance  from  a  line  through  a  focus  of  an  hyperbola,  perpen- 
dicular to  an  asymptote,  to  the  center  is  numerically  equal  to  a. 

6.  The  product  of  the  distances  from  the  asymptotes  to  any  point  on  the 
hyperbola  is  constant. 

7.  The  focal  radius  of  a  point  Pi(xi,  y{)  on  the  parabola  2/2  =  2px  is 


194  ANALYTIC  GEOMETRY 

8.  The  focal  radii  of  a  point  Pi  (xi,  yi)  on  the  ellipse  b^x^  +  a^y^  =  a^b^ 
are  r  =  a  —  exi  and  r'  =  a  -\-  exi. 

9.  The  focal  radii  of  a  point  on  the  hyperbola  b^x^  —  a^y^  =  a'^b^  are 
r  =  exi  —  a  and  r'  =  exi  +  a  when  Pi  is  on  the  right-hand  branch,  or 
r  =  —  exi  —  a  and  r'  =  —  exi  +  a  when  Pi  is  on  the  left-hand  branch. 

10.  The  distance  from  a  point  on  an  equilateral  hyperbola  to  the  center 
is  a  mean  proportional  between  the  focal  radii  of  the  point. 

11.  The  eccentricity  of  an  hyperbola  equals  the  secant  of  the  inclination 
of  one  asymptote. 

80.  Types  of  loci  of  equations  of  the  second  degree.     All  of 

the  equations  of  the  conic  sections  that  we  have  considered  are 
of  the  second  degree.  If  the  axes  be  moved  in  any  manner,  the 
equation  will  still  be  of  the  second  degree  (Theorem  IV,  p.  164), 
although  its  form  may  be  altered  considerably.  We  have  now  to 
consider  the  different  possible  forms  of  loci  of  equations  of  the 
second  degree. 

By  Theorem  VI,  p.  169,  the  term  in  xy  may  be  removed  by 
rotating  the  axes.  Hence  we  only  need  to  consider  an  equation 
of  the  form 

(1)  Ax""  -\-Cif-\-Dx  +  Ey  +  F  =  0. 

It  is  necessary  to  distinguish  two  cases. 
Case     I.    Neither  A  nor  C  is  zero. 
Case  II.    Either  ^  or  C  is  zero. 

A  and  C  cannot  both  be  zero,  as  then  (1)  would  not  be  of  the  second  degree. 

Case  I 

When  neither  A  nor  C  is  zero,  then  ^  =  B^  —  4,  AC  is  not  zero, 
and  hence  (Theorem  VII,  p.  170)  we  can  remove  the  terms  in 
X  and  y  by  translating  the  axes.  Then  (1)  becomes  (Corollary  I, 
p.  171) 

(2)  Ax"" -{- Cy'^ -{- F' =  0. 

We  distinguish  two  types  of  loci  according  as  A  and  C  have  the 
same  or  different  signs. 


CONIC  SECTIONS 


195 


Elliptic  type,  A  and  C  have  the 
same  sign. 

1.  F'  7!^0*    Then    (2)    may    be 

iC2         y2 

written  1 =  1, 

a       /3 

F'  F' 

where    a  = »  /3  = 

A  C 

Hence,  if  the  sign  of  F'  is  different 
from  that  of  A  and  C,  the  locus  is  an 
ellipse;  but  if  the  sign  of  F'  is  the 
same  as  that  of  A  and  C,  there  is  no 
locus. 

2.  F'  =  0.  The  locus  is  a  point. 
It  may  be  regarded  as  an  ellipse 
whose  axes  are  zero  and  it  is  called 
a  degenerate  ellipse. 


Hyperbolic  type,  A  and  C  have  dif- 
ferent signs. 

1.  F'  ^0.*    Then    (2)    may    be 

written  h  —  =  1, 

a      p 

F'  F' 

where    a  = ,  8  = 

A^  C 

Hence  the  locus  is  an  hyperbola  whose 
foci  are  on  the  F-axis  if  the  signs  of 
F'  and  A  are  the  same,  or  on  the 
X-axis  if  the  signs  of  F'  and  C  are 
the  same. 

2.  F'  =  0.  The  locus  is  a  pair  of 
intersecting  lines.  It  may  be  regarded 
as  an  hyperbola  whose  axes  are  zero 
and  it  is  called  a  degenerate  hyperbola. 


Case  II 

When  either  A  or  C  is  zero  the  locus  is  said  to  belong  to  the 
parabolic  type.  We  can  always  suppose  ^1  =  0  and  C  9^  0,  so  that 
(1)  becomes  ^    - 

(3)  Ci/  +  Dx-\-  Ey  +  F=0. 

For  if  ^  5^  0  and  C  =  0,  (1)  becomes  Ax"^  +  Dx  +  Ey  -\- F=(i.   Rotate  the  axes 
It 
(Theorem  II,  p.  162)  through  —  by  setting  x  =  —y',y~  x'.    This  equation  becomes 

Ay'^  +  Ex'  —  Dtf  -\-  F=Q),  which  is  of  the  form  (3). 

By  translating  the  axes  (3)  may  be  reduced  to  one  of  the  forms 

(4)  Cy^-\-Dx  =  0  ox 

(5)  .  Cif  +  F'  =0. 

For  substitute  in  (3),  x  =  x'  +  h,  y  =  y'  -\-  k. 

This  gives 

(6)  Cy'^^-Dx' -^2Ck\y' ^-Ck'^    =0. 

-\-  E      I      -\-Dh 

+  Ek 
+  F 
If  we  determine  h  and  k  from 

2Ck  +  E=0,     Ck^  +  Dh -\- Ek  +  F  =  0, 

[then  (6)  reduces  to  (4).    But  if  Z>  =:  0,  we  cannot  solve  the  last  equation  for  h,  so 

[that  we  cannot  always  remove  the  constant  terra.    In  this  case  (6)  reduces  to  (5). 

*  Read  "  F'  not  equal  to  zero  "  or  "F^  diiferent  from  zero." 


196  ANALYTIC  GEOMETRY 

Comparing  (4)  with  (III),  p.  179,  the  locus  is  seen  to  be  ^.parab- 

I    ¥' 

ola.     The  locus  of  (5)  is  the  pair  of  parallel  lines  y  =  ±^  —  — 

when  F'  and  C  have  different  signs,  or  the  single  line  y  =  0 
when  F'  =  0.     li  F'  and  C  have  the  same  sign,  there  is  no  locus. 
When  the  locus  of  an  equation  of  the  second  degree  is  a  pair  of 
parallel  lines  or  a  single  line  it  is  called  a  degenerate  parabola. 
We  have  thus  proved 

Theorem  XIII.  The  locus  of  an  equation  of  the  second  degree  is 
a  conic,  a  point,  or  a  pair  of  straight  lines,  ivhich  may  he  coincident. 
By  moving  the  axes  its  equation  m^ay  he  reduced  to  one  of  the  three 

forms  .^ 

Ax"  +  Cif  +  F*  =  0,   Cy'^  -^  Dx  =  0,   Cy^  +  F*  =  0,  '* 

where  A ,  C,  and  D  are  different  from  zero. 

Corollary.    TJie  locus  of  an  equation  in  which  the  term  in  xy  is 

lacking,  ^^^2  _^  ^f  +  Dx  +  Ey  +  F  =  0, 

will  helong  to 

the  parabolic  type  if  A  =  0  or  C  =  0, 

the  elliptic  type  if  A  and  C  have  the  same  sign, 

the  hyperbolic  type  if  A  and  C  have  different  signs. 

PROBLEMS 

1 .  To  what  point  is  the  origin  moved  to  transform  (1)  into  (2)  ?  - 


( -  —  -— ^ 

V      2A'       20/ 


Ans.    . 

V      2A        2C. 

2.  To  what  point  is  the  origin  moved  to  transform  (3)  into  (4)  ?   into  (5)  ? 

3.  Simplify  Ax^  +  Dx-\-Ey  +  F=Ohy  translating  the  axes  (a)  if  ^  7^  0, 
(b)  it  E  =  0,  and  find  the  point  to  which  the  origin  is  moved. 


(b)  Jx2  +  F' =  0,   (-^'O) 


2A 

*  In  describing  the  final  form  of  the  equation  it  is  unnecessary  to  indicate  by  primes 
what  terms  are  different  from  those  in  (1). 


CONIC  SECTIONS  197 

4.  To  what  types  do  the  loci  of  the  following  equations  belong? 

(a)  4:X^  +  y^  -  ISx  -{-  7  y  -  1  =  0.         (e)  x^  +  1  y^  -  8x  +  1  =  0. 

(b)  2/2  +  3»  -  4y  +  9  =  0.  (f)  x^  +  y2-6x  +  8y  =  0. 

(c)  121a;2-44y2  +  68x-4  =  0.  (g)  3x2-42/2  -62/  + 9  =  0. 

(d)  x2  +  4?/-3  =  0.  (h)  x2-8x  + 9?/ -11  =  0. 

(i)  The  equations  in  problem  1,  p.  172,  which  do  not  contain  the  xy-term. 

81.  Construction  of  the  locus  of  an  equation  of  the  second 
degree.    To  remove  the  xi/-teim.  from 

(1)  Ax^  +  Bxi/  -\-  Cy-^Dx-\-  Ey-\-  F=0 

it  is  necessary  to  rotate  the  axes  through  an  angle  6  such  that 
(Theorem  VI,  p.  169) 

(2)  tan2^  =  j4^, 

while  in  the  formulas  for  rotating  the  axes  [(II),  p.  162]  we  need 
sin  6  and  cos  $.     By  1  and  3,  p.  19,  we  have 

(3)  cos  2  ^  =  ± 


Vl  -f-  tan^  2  0 


From  (2)  we  can  choose  2  ^  in  the  first  or  second  quadrant  so 
the  siffn  in  (3)  must  be  the  same  as  in  (2).  0  will  then  be  acute; 
and  from  15,  p.  20,  we  have 


/I -cos  2^          ^        ,   ^/l  +  cos2^ 
(4)  sm  ^  =  +  \j ,  cos  6>  =  +  V 

In  simplifying  a  numerical  equation  of  the  form  (1)  the  com- 
putation is  simplified,  if  A  =  B^  —  4:AC=^0,  by  first  removing 
the  terms  in  x  and  ?/  (Theorem  VII,  p.  170)  and  then  the  a^y-term. 

Hence  we  have  the 

Rule  to  construct  the  locus  of  a  numerical  equation  of  the  second 
degree. 

First  step.    Compute  A  =  B^  —  4^C. 
Second  step.    Simplify  the  equation  by 

(a)  translating  and  then  rotating  the  axes  if  ^^  0', 

(b)  rotating  and  then  translating  the  axes  if  A  =  0. 


198  ANALYTIC  GEOMETRY 

Third  step.  Determine  the  nature  of  the  locus  hy  inspection  of 
the  equation  (§  80,  p.  194). 

Fourth  step.    Plot  all  of  the  axes  used  and  the  locu^. 

In  the  second  step  the  equations  for  rotating  the  axes  are 
•found  from  equations  (2),  (3),  (4),  and  (11)^  p.  162.  But  if  the 
xy-tQim.  is  lacking,  it  is  not  necessary  to  rotate  the  axes.  The 
equations  for  translating  the  axes  are  found  by  the  Eule  on 
p.  165. 

Ex.  1.    Construct  and  discuss  the  locus  of 

x2  +  4  x?/  +  4  y2  4.  12  X  -  6  2/  =  0. 

Solution.    First  step.    Here  A  =  42-4-l-4  =  0. 

Second  step.    Hence  we  rotate  the  axes  through  an  angle  d  such  that, 

'''''^'      ■  4  4,, 

tan2^  =  — = \? 

1-4  3 

Then  by  (3),  cos2^  =  -|, 

2                            1 
and  by  (4),  sin^  =  — —  and  cos^  — 

V5  Vs 


(1) 


The  equations  for  rotating  the  axes  [{II),  p.  162]  become 

x'  —  2y'  Ix'  -\-y' 

^  = rr-,  y  = 

V5  V5 

Substituting  in  the  given  equation,*  we  obtain 


a;'2 -2/'  =  0. 

V5 

It  is  not  necessary  to  translate  the  axes. 
Third  step.    This  equation  may  be  written 

x"^  =  -^y'. 

3 
Hence  the  locus  is  a  parabola  for  which  p  =  — z:,  and  whose  focus  is  on 

the  F'-axis.  "^^ 

*  When  A  =  0  the  terms  of  the  second  degree  form  a  perfect  square.    The  work  of 
substitution  is  simplified  if  the  given  equation  is  first  written  in  the  form 

(x  +  2y)^  +  12x-(iy  =  0. 
It  will  be  shown  in  Chapter  XII  that  when  A  =  0  the  locus  is  always  of  the  parabolic 
type. 


CONIC  SECTIONS 


199 


Fourth  step.    The  figure  shows  both  sets  of  axes,*  the  parabola,  its  focus 
and  directrix. 

In  the  new  coordinates  the  focus 

is  the  point  (  0,  — ^  )  and  the  direc- 
^      2V5^ 

Q 

trix  is  the  line  y'  = (Theorem 

2V5 
The  old  coordinates  of 


:^ 

f^w^ 

V  _ 

N 

^. 

.     Jt 

!v. 

V  2t 

XJt  ' 

""^^^ 

7~t     - 

^^U 

"tt^^  -' 

t     ^^. 

IV,  p.  179). 

the  focus  may  be  found  by  substi- 
tuting the  new  coordinates  for  x' 
and  y'  in  (1),  and  the  equation  of 
the  directrix  in  the  old  coordinates 
may  be  found  by  solving  (1)  for  y' 
and  substituting  in  the  equation  given  above. 

Ex.  2.    Construct  the  locus  of 

5  a;2  +  6  xy  +  5  2/2  +  22  (c  -  6  y  +  21  =  0. 
Solution.    First  step.    A  =  6^-4'6-S7^0. 

Second  step.    Hence  we  translate  the  axes  first.    It  is  found  that  the  equa- 
tions for  translating  the  axes  are 

X  =  x'  -  i,  y  =  y'  -\-  S, 
and  that  the  transformed  equation  is 

5x'^  +  6  xY  +  5  /2  ^  32. 
From  (2)  it  is  seen  that  the  axes  must  be  rotated  through  — .     Hence  we 


set 


Y 

N 

1 

Y 

Y 

V^ 

\l. 

/ 

j 

\ 

^ 

s, 

/ 

1 

\ 

>< 

V 

^ 

\, 

\ 

^ 

/ 

> 

>< 

\ 

z 

\ 

^ 

\ 

' 

— 

y 

z 

— 

\ 

0 

•'  y  = 


X"  +  y" 


V2  V2 

and  the  final  equation  is 

4.x'"^  +  y'"^  =  \Q. 

Third  step.    The  simplified  equa- 
tion may  be  written 


16 


1. 


Hence  the  locus  is  an  ellipse  whose  major  axis  is  8,  whose  minor  axis  is  4, 
and  whose  foci  are  on  the  F''-axis. 

Fourth  step.    The  figure  shows  the  three  sets  of  axes  and  the  ellipse. 


*The  inclination  of  OX'  is  0,  and  hence  its  slope,  tan  6,  may  be  obtained  from  (4). 
2     .     1 

method  given  in  the  footnote,  p.  35. 


^1  ■      •          ,     ,       „     sin  0 
this  example  tan  6  = 


In 


=  2,  and  the  X'-axis  may  be  constructed  by  the 


200  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Simplify  the  following  equations  and  construct  their  loci,  foci,  and 
directrices. 

(a)  3x2  -4x2/  +  8x-l=0.  Ans.    x'"^  -  4  y'"^  +  1  =  0. 

(b)  4x2  +  4 x?/  +  2/2  +  8 X  -  16 ?/  =  0.         Anz.    5 x'2  -  8  V5 y'  =  0. 

(c)  41  x2  -  24  X2/  +  34  ?/  +  25  =:  0.  Ans.    x'2  +  2  ?/'2  _^  1  ^  0. 

(d)  17x2 -12x?/  +  8?/2-68x  + 24?/ -12  =  0. 

An&.  x''2  +  4  2/"2  _  16  =  0. 

(e)  2/2  +  6  X  -  6  y  +  21  =  0.  Ans.  y"^  +  6  x'  =  0. 

(f )  x2  -  6  X2/  +  9  2/2  +  4  X  -  12  ?/  +  4  =  0.  Ans.  y'"^  =  0. 

(g)  12xy  -  52/2  +  48?/  -  36  =  0.  Ans.  4x''2  _  g^/'^a  =  3(3. 
(h)  4x2  _l2xy +  92/2  + 2x- 3  2/ -12  =  0. 

Ans.    52  2/''2  -  49  =  0. 
(i)  14x2  -  4x2/ +  112/2 -88x  + 34?/ +  149  =  0. 

Ans.    2x''^  +  Sy''^  =  0. 
(j)  12x2  +  8x2/ +  18  2/2 +  48  x  + 16  2/  + 43  =  0. 

^ns.    4x2  +  22/2  =  1. 
(k)  9x2  +  24x2/ +  162/2 -36x- 482/ +  61  =  0. 

^ns.    x"2  +  1  =  0. 
(1)  7  x2  +  50  X2/  +  7  2/2  =  50.  Ans.    16  x'2  -  9  2/^2  =  25. 

(m)  x2  +  3X2/  -  32/2  +  6x  +  92/  +  9  =  0.    ^ns.    3x''2  -  7 2/^2  ^  q. 
(n)  16x2  -  24x2/  +  O?/^  -  60x  -  80 2/  +  400  =  0. 

Ans.    y'"^  -  4  x''  =  0. 
(o)  95x2  +  56a;y  _  IO2/2  -  56x  +  2O2/  +  194  =  0. 

^MS.   Qx"'^-y"-'^^Vl  =  Q. 
(p)  5  x2  -  5  xy  -  7  2/2  -  165  x  +  1320  =  0.  Ans.    15  x''2  _  \\  y"%  _  330  =  0. 

82.  Systems  of  conies.  The  purpose  of  this  section  is  to 
illustrate  by  examples  and  problems  the  relations  between 
conies  and  degenerate  conies  and  between  conies  of  different 
types. 

A  system  of  conies  of  the  same  type  shows  how  the  degenerate 
conies  appear  as  limiting  forms,  while  a  system  of  conies  of  dif- 
ferent types  shows  that  the  parabolic  type  is  intermediate  between 
the  elliptic  and  hyperbolic  types. 

Ex.  1.    Discuss  the  system  of  conies  represented  by  x2  +  4  2/2  =  k. 

Solution.  Since  the  coefficients  of  x2  and  y^  have  the  same  sign,  the  locus 
belongs  to  the  elliptic  type  (Corollary,  p.  196).  When  k  is  positive  the  locus 
is  an  ellipse ;  when  fc  =  0  the  locus  is  the  origin,  —  a  degenerate  ellipse  /  and 
when  k  is  negative  there  is  no  locus. 


CONIC  SECTIONS 


201 


In  the  figure  the  locus  is  plotted  for  k  =  100,  64,  36,  16,  4,  1,  0.  It  is  seen 
that  as  k  approaches  zero  the  ellipses  become  smaller  and  finally  degenerate 
into  a  point.     As  soon  as  k  becomes  negative  there  is  no  locus.     Hence  the 


point  is  a  limiting  case  between  the  cases  when  the  locus  is  an  ellipse  and 
when  there  is  no  locus. 

Ex.  2.    Discuss  the  system  of  conies  represented  by  4  x^  —  16  y^  =  k. 
Solution.    Since  the  coefficients  of  x^  and  y^  have  opposite  signs,  the  locus 


-3 

L^ 
k 

^ 

■^ 

— 

- 



Y_ 

— 

— 

_ 

— 

— 

— 

— 

^ 

^ 
^ 

V 

^ 

N 

^ 

s 

^ 

*«», 

^ 

•^ 

^ 

^ 

^ 

^ 

:> 

*^; 

s 

^ 

1 

^ 

^ 

-*- 

k- 

-21 

6^ 

$ 

^ 

^ 

^* 

\ 

S 

^ 

^ 

frs 

^ 

/ 

»"^ 

N 

N 

^ 

^ 

>s 

..^ 

k= 

-6 

^^ 

yfi 

^ 

/ 

A 

\ 

\ 

N 

v^ 

^ 

>s 

1:/-^ 

^ 

^^\/ 

^^ 

/ 

V 

\ 

\ 

\ 

v 

"S 

^ 

\^^ 

j/\ 

(^ 

^     / 

f 

A" 

J 

/ 

/ 

x* 

^ 

^ 

k 

V 

I 

/ 

/ 

/ 

X 

^ 

>' 

:^ 

«> 

S 

\ 

\ 

/ 

y 

/ 

< 

^ 

^ 

•^ 

—J 

<^ 

:^ 

s 

N 

^^ 

/ 

/ 

> 

-ii 

^ 

^ 

^ 

^ 

=^^ 

\ 

^ 

.^ 

^ 

^ 

^ 

^ 

r 

•^ 

^ 

^ 

% 

;5: 

s, 

y' 

i 

t- 

^ 

z^ 

^ 

-^ 

v; 

^ 

^ 

s, 

.^ 

^ 

--' 

**^ 

V 

^ 

^ 

H 

r' 

1 

S*J 

belongs  to  the  hyperbolic  type.     The   hyperbolas  will  all  have  the  same 
asymptotes  (p.  189),  namely,  the  lines  cc  ±  2  y  =  0.    The  given  equation  may 


be  written 


k       k 


4       16 

The  locus  is  an  hyperbola  whose  foci  are  on  the  X-axis  when  k  is  positive  and 


202 


ANALYTIC  GEOMETRY 


on  the  F-axis  when  k  is  negative.     For  A:  =  0  the  given  equation  shovs^s  that 
the  locus  is  the  pair  of  asymptotes. 

In  the  figure  the  locus  is  plotted  for  k  =  256,  144,  64, 16,  0,  -  64,  -  256. 
It  is  seen  that  as  k  approaches  zero,  whether  it  is  positive  or  negative,  the 
hyperbolas  become  more  pointed  and  lie  closer  to  the  asymptotes  and  finally 
degenerate  into  the  asymptotes.  Hence  a  pair  of  intersecting  lines  is  a  lim- 
iting case  between  the  cases  when  the  hyperbolas  have  their  foci  on  the  X-axis 
and  on  the  Y-axis. 

Ex.  3.    Discuss  the  system  of  conies  represented  hj  y'^  =  2kx  ■\- 16. 

Solution.  As  only  one  term  of  the  second  degree  is  present,  the  locus 
belongs  to  the  parabolic  type  (Corollary,  p.  196).    The  given  equation  may  be 

simplified  (Rule,  p.  165)  by  translating  the  axes  to  the  new  origin  ( ,  0  Y 

We  thus  obtain  \     k      y 

Q 

The  locus  is  therefore  a  parabola  whose  vertex  is  ( ,  0)  and  for  which 

p  =  k.    It  will  be  turned  to  the  right  when  k  is  positive,  and  to  the  left  when 
k  is  negative.     But  if  A:  =  0,  the  locus  is  the  degenerate  parabola  y  =  ±  4. 


^ 

Y 

^ 

-^ 

'■**s, 

V, 

^ 

^ 

^ 

'-' 

^ 

" 

■■ 

-— 

- 

"^ 

N 

s. 

y 

y 

^ 

^ 

, 

- 

- 

^— 

^ 

"*^ 

«^ 

\ 

/" 

-^ 

^ 

— 

" 

■» 

. — 

- 

k^l> 

■^ 

a* 

^ 

k 

= 

^ 

— " 

"^ 

^ 

:^ 

=? 

^ 

^ 

V 

-^ 

^ 

/ 

7 

\ 

"^ 

■^ 

^- 

■#■ 

^ 

\v 

y 

/ 

/ 

Y 

■t 

\ 

N 

s< 

X 

K 

r 

y 

•^ 

f 

M 

Vl 

to 

\ 

> 

V; 

- 

JT 

^ 

i 

0 

i 

j 

X 

V. 

'^ 

\ 

^ 

\j 

y 

/ 

y 

^ 

^ 

^ 

^ 

:^ 

\ 

\ 

/ 

^ 

^ 

^^ 

^ 

kH 

1 

^ 

^ 

^ 

'" 

, 

^ 

ff» 

^ 

^ 

^ 

c=: 



^ 

- 

- 

— 

P 

^ 

r-^ 

/ 

y 

\ 

^^ 

*^ 

^ 

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— 

. 

. 

"^ 

^ 

y 

s 

s 

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^ 

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— 

-^ 

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e 

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r 

M 

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■ 

In  the  figure  the  locus  is  plotted  for  A;  =  ±  4,  ±  2,  ±  1,  db  |,  0.  It  is  seen 
that  as  k  approaches  zero,  whether  it  is  positive  or  negative,  the  vertex  recedes 
from  the  origin  and  the  parabola  lies  closer  to  the  lines  ?/  =  ±  4  and  finally 
degenerates  into  these  lines.  The  degenerate  parabola  consisting  of  two 
parallel  lines  appears  as  a  limiting  case  between  the  cases  when  the  parab- 
olas are  turned  to  the  right  and  to  the  left. 


CONIC  SECTIONS 


203 


Ex.  4.    Discuss  the  system  represented  by 
Solution. 


x2 


25  -  A;  9  -  A: 
When  fc  <  9  the  locus  is  an  ellipse  whose  foci  are  {±  c,  0)  where 
c2  =  (25  -k)  -  {d  -k)  =  16  (theorem  V,  p.  185).  When  9  <  fc  <  25  the  locus 
is  an  hyperbola  whose  foci  are  (±  c,  0),  where  c^  =  (25  -  fc)  -  (9  -  fc)  =  10 
(Theorem  VI,  p.  185).  When  A;  >  25  there  is  no  locus.  Since  the  ellipses  and 
hyperbolas  have  the  same  foci,  (±  4,  0),  they  are  called  confocal. 
Clearing  of  fractions,  we  obtain 

(9  -  A:)x2  +  (25  -k)y^  =  {9-  k)  (25  -  k).       . 

Hence  when  A;  =  9  or  25  the  locus  is  a  degenerate  parabola  y^-OoTX^  =  0. 

In  the  figure  the  locus  is  plotted  for  k  =  -  56,  -  24,  0,  7,  9,  11,  16,  21, 

24,  25.     As  k  increases  and  approaches  9  the  ellipses  flatten  out  and  fmally 


1 1 1 1 1 1\[  1  l\J  l: 

\i\f\     1/          1 — 

N       )<^\'T~ 

"^E^2        ^^ 

^\  •^'^     I *=■ 

2^   /      '  ^    y 

X     S^P~ 

~H-2     5 

/  v/"!  \ 

i  2^/  s^ 

^^»J          Z  S,      tjL'^  = 

oil  A  \^^ 

/  "^^^C         s.^'x      7 

^t$^^  \-"% 

t  t'^^%^--- 

-~ty\.^%  4 

-t-  ^t- 

Cu          ^'-^  ^ 

^         IT    X^     " 

_F^y                                    ^ 

,    L'^NZl-- 

-qrt"^7  i 

5,"^^    ?> 

\^\  \  t>ti 

-"''  V    5       L.J 

X^  ^Z  -t~'^^^ 

^^^^5    t 

X  S^\=z^ 

^^^    ^■^^- 

-^"^  ^^i 

^     ^S  -^ 

JL  s^   ^^ 

■'^       /^U^ 

-fc-'  N'^          ^ 

-  ==  r--i  ,. 

E    ^    1 

-degenerate  into  the  JT-axis,  and  as  k  decreases  and  approaches  9  the  hyper- 
bolas flatten  out  and  degenerate  into  the  X-axis.  Hence  the  locus  of  the 
parabolic  type,  y'^  =  0,  appears  as  a  limiting  case  between  the  ellipses  and 
hyperbolas.  As  k  increases  and  approaches  25. the  two  branches  of  the 
hyperbolas  lie  closer  to  the  F-axis,  and  in  the  limit  they  coincide  with 
the  F-axis. 

Ex.  5.    Plot  and  discuss  the  locus  of  A:x2  +  2  2/2  -  8  x  =  0. 

Solution.    If  A:  =  0,  the  locus  is  a  parabola.     If  k  is  not  zero,  the  locus  is 


an  ellipse  or  hyperbola  according  as  k  is  positive  or  negative, 
passes  through  the  origin  for  all  values  of  k. 


The  locus 


204 


ANALYTIC  GEOMETRY 


Simplifying  by  translating  the  axes  (Rule,  p.  165),  it  is  found  that  if  the 
origin  is  (  - ,  0  j  the  equation  becomes 


From  this  the  axes  may  be  determined  and  the  locus  sketched. 

In  the  figure  the  locus  is  plotted  for  fc  =  1,  f,  ^,  0,  —  1,  —  i.    If  A;  is  posi- 
tive and  approaches  zero,  the  ellipses  become  longer  and  lie  closer  to  the 


\ 

Y\ 

, 

y^ 

S 

^ 

ft 

y 

■b- 

^ 

•v 

^ 

s. 

4- 

A 

f. 

P 

y 

"^ 

V 

N 

^^ 

y 

y^ 

\^ 

=s  ( 

-^ 

' 

V 

s, 

^• 

\ 

^ 

y 

^ 

S 

^>' 

\ 

\ 

A 

^ 

^ 

N 

y 

J 

^ 

> 

7: 

<" 

\ 

r 

^t 

\ 

f»» 

\ 

J 

' 

) 

0 

\ 

} 

) 

) 

X 

/ 

^ 

> 

^ 

^ 

^ 

^ 

^ 

y 

,/ 

y 

^ 

s> 

y 

/^' 

y 

% 

N 

"^ 



^ 

X 

/ 

/ 

N 

^ 

■"v 

^ 

- 

>• 

y 

/ 

's 

V 

V 

- 

y 

X 

^ 

y 

^ 

y' 

parabola.  If  k  is  negative  and  approaches  zero,  the  right-hand  branches  of 
the  hyperbolas  lie  closer  to  the  parabola  and  the  left-hand  branches  recede 
from  the  origin.  This  shows  that  tUe  parabola  is  a  limiting  form  between  the 
ellipse  and  hyperbola. 

How  does  the  locus  behave  if  k  approaches  +  co  or  —  co  ?  >- 


PROBLEMS 

1.  Plot  on  separate  sheets  the  foci  and  directrices  of  the  conies  plotted  in 
examples  1,  2,  and  3.  Where  are  the  foci  and  directrices  Of  the  degenerate 
conic  in  each  system  ?     Verify  the  results  analytically. 

2.  Plot  the  following  systems  of  conies  and  show  that  the  conies  of  each 
system  belong  to  tlie  same  type.  Draw  enough  conies  so  that  the  degenerate 
conies  of  the  system  appear  as  limiting  cases. 


<^>S+ 


k. 


(c) 


16 


=  k. 


(b)  2/2  =  2  kx.  (d)  x2 

3.  Problem  1  for  the  systems  in  problem  2. 


r 

9 
2ky  -6. 


CONIC  SECTIONS  205 

4.  Plot  the  system  —  +  —  =  1  for  positive  values  of  k.    What  is  the  locus 

k      16 

if  k  =  lQ?  Show  how  the  foci  and  direatrices  behave  as  k  increases  or 
decreases  and  approaches  16.  Where  are  the  foci  and  directrices  of  a 
circle  ? 

5 .  Plot  the  system  in  problem  4  for  positive  and  negative  values  of  k.  Show 
hov^the  conies  change  as  k  approaches  zero  when  it  is  positive  and  negative. 

6.  Plot  the  following  systems  of  conies  and  show  that  all  of  the  conies  of 
each  system  are  confocal.  Discuss  degenerate  cases  and  show  that  two 
conies  of  each  system  pass  through  every  point  in  the  plane. 

^  '  16-A;      SQ-k  ^  '  64  -  k      16  -  k 

(b)  y^  =  2kx-}-  A;2.  (d)  x^  =  2ky  +  k^. 

7.  Plot  and  discuss  the  systems 

(a)  16  {X  _  A;)2  +  9  2/2  =  144.  (c)  {y  -  k)"^  =  4  x. 

(b)  xy  =  k.  (d)  4 (X  -  k)'^  -9{y-  ky  =  36. 

8.  Plot  the  following  systems  and  discuss  the  locus  as  k  approaches  zero 
and  infinity.     Show  how  the  foci  and  directrices  behave  in  each  case. 

^    (^-^2        .^  {x-kl_y^^ 

^  '        k^  36  ^  ^        A;2  36 

9.  Show  that  all  of  the  conies  of  the  following  systems  pass  through  the 
points  of  intersection  of  the  conies  obtained  by  setting  the  parentheses  equal 
to  zero.     Plot  the  systems  and  discuss  the  loci  for  the  values  of  k  indicated. 

(a)  (?/2-4x)  +  fc(z/2  +  4rc)  =  0,  fc3z:+l,  -1. 

(b)  {x^  +  ?/2  _  16)  +  fc(x2  -  2/2  _  4)  3z:  0,  A;  =  +  1,  -  1,  -  4. 

(c)  (x2  +  2/2  -  16)  +  fc(a;2  -  2/2  -  16)  =  0,  fc  =  +  1,  -  1. 

(d)  (x2  +  16  2,2  -  64)  +  A:  (x2  -  4  2/2  _  36)  =  0,k  =  -l,4,  -  -y. 

(e)  x2  +  4  2/  +  A:  (x2  -  4  y  +  16)  =  0,  A;  =  +  1,  -  1. 

MISCELLANEOUS   PROBLEMS 

1.  Construct  the  loci  of  the  following  equations,  their  foci  and  directrices. 

(a)  9x2 +  24x2/+ 162/2- 50x4-802/ -275  =  0. 

(b)  56x2-64x2/  + W)9y2._i76x  + 282 y- 896  =  0. 

(c)  5x2  _  I2xy  +  6x  -  S6y  -  63  =  0. 

2.  Find  the  value  of  p  if  ^2  =  2px  passes  through  the  point  (3,  -  1). 

X2         2/2 

3.  Find  the  values  of  a  and  h  if  —  +  ^  =  1  passes  through  the  points 
(3,  -  6)  and  (4,  8).  "^      ^^ 

4.  Find  the  equation  of  the  locus  of  a  point  P  if  the  sum  of  its  distances 
from  the  points  (c,  0)  and  (—  c,  0)  is  2  a, 


206  ANALYTIC  GEOMETRY 

6.  Find  the  equation  of  the  locus  of  a  point  P  if  the  difference  of  its 
distances  from  the  points  (c,  0)  and  ( —  c,  0)  is  2  a. 

6.  Find  the  equation  of  the  locus  of  a  point  if  its  distances  from  the  line 
x  =  —  —  and  the  point  (  — ,  0  j  are  equal. 

7.  Show  that  a  conic  or  degenerate  conic  may  be  found  which  satisfies 
five  conditions,  and  formulate  a  rule  by  which  to  find  its  equation. 

Hint.  Compare  p.  93  and  p.  133. 

8.  Find  the  equation  of  the  conies  which  satisfy  the  following  conditions. 

(a)  Passing  through  (0,  0),  (1,  2),  (1,  -  2),  (4,  4),  (4,  -  4). 

(b)  Passing  through  (0,  0),  (0,  1),  (2,  4),  (0,  4),  (-  1,  -  2). 

(c)  Passing  through  (3,  7),  (4,  6),  (5,  3)  if  ^  =  -B  and  C  =  0. 

(d)  Passing  through  (1,  2),  (3,  4),  {4,  2),  (2,  -  1),  (4,  2). 

(e)  Passing  through  (0,  0),  (0,  1),  (1,  0),  (6,  6),  (5,  6). 

(f)  Passing  through  (0,  0),  (2,  0),  (-  3,  2),  (5,  2)  with  its  axes  parallel  to 
the  coordinate  axes. 

9.  What  is  the  nature  of  a  conic  which  passes  through  five  points,  of 
which  three  or  four  are  on  a  straight  line  ? 

The  circle  whose  radius  is  a  and  whose  center  is  the  center  of 
a  central  conic  is  called  the  auxiliary  circle. 

10.  The  ordinates  of  points  on  an  ellipse  and  the  auxiliary  circle  which 
have  the  same  abscissas  are  in  the  ratio  of  6  :  a, 

11.  The  area  of  an  ellipse  is  Ttab. 

Hint.  Divide  the  major  axis  into  equal  parts.  With  these  as  hases  inscribe  rectan- 
gles in  the  ellipse  and  auxiliary  circle.  Apply  problem  10  and  increase  the  number  of 
rectangles  indefinitely. 

12.  The  auxiliary  circle  of  an  hyperbola  passes  through  the  intersections 
of  the  directrices  and  asymptotes. 

13.  Show  that  the  locus  of  xy  -^  Dx  ■}-  Ey  -\-  F  =  0  is  either  an  equilateral 
hyperbola  whose  asymptotes  are  parallel  to  the  coordinate  axes  or  a  pair  of 
perpendicular  lines. 

14.  Discuss  the  form  of  the  locus  oi  x^  -  y^ -\-  Dx -{-  Ey  +  F  =  0. 


CHAPTER  IX 
TANGENTS   AND   NORMALS 

83.  The  slope  of  the  tangent.  Let  Pi  be  a  fixed  point  on  a 
curve  C  and  let  P^  be  a  second  point  on  C  near  P^.  Let  P^ 
approach  P^  by  moving  along  C.  Then  the  limiting  position 
PiT  of  the  secant  through  Pi  and  P^  is  called  the  tangent  to  C 
at  Pi. 

It  is  evident  that  the  slope  of  PiT  is  the  limit  of  the  slope 
of  P1P2.     The  coordinates  of  Pg  ^^y  be  written  (xi  +  A,  2/1  +  k), 


where  h  and  k  will  be  positive  or  negative  numbers  according 
to  the  relative  positions  of  Pi  and  Pg.  The  slope  of  the  secant 
through  Pi  and  Pg  is  therefore  (Theorem  V,  p.  35) 

3/1  -fa  -^k^     k 


(1) 


Xi  —  Xi 


h)     h 


As  P2  approaches  Pi  both  h  and  k  approach  zero,  and  hence 
0 


k 


approaches  -?  which  is  indeterminate.    The  actual  value  of  the 

k 
limit  of  -  may  be  found  in  any  case  from  the  conditions  that' 

Pi   and   P2   lie   on   C   (Corollary,   p.   53),   as   in   the   example 
following. 

207 


208 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  slope  of  the  tangent  to  the  curve  C  :  8  y'=  x^  at  any  point 
Pi  {xu  yi)  on  G. 

Solution.    Let  Pi  (xi,  yi)  and  P^  (xi  +  ^,  2/i  +  k)  be  two  points  on  C. 
Then  (Corollary,  p.  53) 

(2)  8  2/i  =  Xi3 

and  8  (^i  +  k)  =  (xi  +  h)\ 

or 

(3)  8  ?/i  +  8  A;  =  Xi3  +  3  XiS^i  +  3  x^h^  +  /i^. 

Subtracting  (2)  from  (3),  we  obtain 

^k  =  ^Xx^h  +  Zxrh'^  +  h^. 
Factoring,     8  A;  =  /i  (3  Xi2  +  3  XiA  +  TjF)  ; 

fc  _  3xi2_+_3xiM-^ 
h~  8 


y\ 

1 

H 

I 

f 

'/ 

^^ 

P^. 

-J 

h 

X 

/^ 

"o 

/ 

X 

/ 

/ 

f 

/ 

1 

1 

* 

/ 

/ 

/ 

Y 

and  hence 


Then,  as  P2  approaches  Pi,  h  and  k  approach 
zero  and  the 

limit  of  -  —  limit  of = 

'  h  8  8  ■ 

3xi2 
Hence  the  slope  m  of  the  tangent  at  Pi  is  m  =  — -• 

8 
C  is  symmetrical  with  respect  to  0,  and  the  tangents  at  symmetrical  points 
are  parallel  since  only  even  powers  of  Xi  and  yi  occur  in  the  value  of  m.     The 
tangent  at  the  origin  is  remarkabL  In  that  it  crosses  the  curve. 

The  method  employed  in  this  example  is  general  and  may  be 
formulated  in  the  following 

Rule  to  determine  the  slope  of  the  tangent  to  a  curve  C  at  a  point 
Pi  on  C. 

First  step.  Let  Pi  (xi,  1/1)  and  P^  {xi  +  h,  yi  +  k)  he  two  points 
on  C.  Substitute  their  coordinates  in  the  equation  of  C  and 
subtract. 

Second  step.  Solve  the  result  of  the  first  step  for  —  ?*  the  slope 
of  the  secant  through  Pi  and  P^. 

Third  step.  Find  the  limit  of  the  result  of  the  second  step  when 
h  and  k  approach  zero.     This  limit  is  the  required  slope. 


*  The  sohxtioii  will  contain  h  and  k  separately,  so  that  the  equation  is  not  solved  in  the 
ordinary  sense. 


TANGENTS  AND  NORMALS 


209 


Ex.  2.  Find  the  slope  of  the  tangent  to  the  semicubical  parabola  3  2/2  =  x^ 
atPi(iCi,  2/i). 

Solution.  First  step.  Let  Pi(iCi,  yi)  and  P^ixi  +  ^,  2/i  +  k)  be  two  points 
on  the  curve.     Then  (Corollary,  p.  53) 

(4)  3?/i2  =  Xi3 

and  3  yi2  +  6  kyi  +  3  A;^  =  xi^  +  3  x^^h  +  3  Xih"^  +  ^8. 
Subtracting, 

6yik  +  3A;2  =  Bxi^h  +  3xi;i2  _i_  ^3, 
Second  step.    Factoring, 

k{6yi  +  Sk)  =  h{Sxi^  +  S  Xih  +  ^^2). 
k  _Sxi^  +  Sxih  +  h^ 
h~ 


Hence 


Qyi  +  Bk 
Third  step.    As  h  and  k  approach  zero, 

limit  of-  =  limit  of- 
h 


62/1 +  3A; 


3xi2 
62/1 


Xj^ 

22/1 


Hence  the  slope  of  the  tangent  at  Pi  is  m  = 

At  the  origin  m  =  -  and  is  indeterminate.     To  find  the  value  of  m  at  the 

origin,  we  may  either  apply  the  rule  a  second  time,  setting  cci  =  0  and  2/1  =  0, 
or  eliminate  2/1  from  the  value  of  m  by  means  of  (4),  thus  obtaining  a  value 
which  is  determinate  at  the  origin. 


PROBLEMS 


1.  Find  the  slopes  of  the  tangents  to  the  following  curves  at  the  points 
indicated. 


(a)  2/2  =  8 X,  Pi  (2,  4). 

(b)  x2  +  2/2  =  25,  Pi  (3,  -4). 

(c)  4x2  +  2/2  =  16,  Pi(0,  4). 

(d)  x2  -  92/2  =  81,  Pi  (15,  -  4). 


Ans.  1. 
Ans.  |. 
Ans.   0. 


2.  Find  the  slopes  of  the  tangents  to  the  following  curves  at  the  point 
Pi(xi,  2/1). 
(a)  2/2  =  6x. 


3 

Ans.   —  • 


(b)  16  2/  =  x4 

(c)  x2  +  2/2  =  16. 


Ans. 


Ans. 


Xi« 


Atis. 

2/1 

Ans. 

3xi2  +  2xi 

2yi 

Ans. 

8-4xi 
2/1-1 

Ans. 

Xi 

A  nQ 

2/1 

./I /to. 

Xi  +  2yj_ 

Ans. 

4-Xi 
2-2/1 

Ans. 

Xi  +  S 

210  ANALYTIC  GEOMETRY 

(d)  x2  -  2/2  =  4. 

(e)  2/2  =  x3  +  x2. 

(f)  4x2  4-2/2-16x-22/  =  0. 

(g)  xy  =  a\ 

(h)  X2/  +  2/2  =  8. 

(i)  x2  -  y2  _  8  X  +  4  2/  =  0. 

(j)  x2  +  2/2  +  6x-82/  =  0. 

4-2/1 

84.  Equations  of  tangent  and  normal.    We  have  at  once  the 

Rule  to  find  the  equation  of  the  tangent  to  a  curve  C  at  a  point 
Pi{^i,  Vi)  on  C. 

First  step.  Find  the  slope  m  of  the  tangent  to  C  at  Pj  {Rule, 
p.  208). 

Second  step.  Substitute  Xi,  yi,  and  m  in  the  point-slope  form  of 
the  equation  of  a  straight  line  [(V),  p.  95]. 

Third  step.  Simplify  that  equation  hy  means  of  the  condition 
that  Pi  lies  on  C  {Corollary,  p.  53). 

Ex.  1.    Find  the  equation  of  the  tangent  to  C  :  8  y  =  x^  at  Pi  (xi,  2/1). 

3xi2 
Solution.    First  step.    From  Ex.  1,  p.  208,  the  slope  is  m  = 

Second  step.    Hence  the  equation  of  the  tangent  is 

2/  -  2/1  =  -x-  {X  -  Xi), 
or  ^ 

(1)  3xi2x  -  8  2/  -  3xi3  +'82/1  =  0. 

Third  step.    Since  Pi  lies  on  C,  8  2/1  =  Xi^. 
Substituting  in  (1),  we  obtain 

(2)  3  Xi2x  -  8  2/  -  2  Xi3  =  0. 

The  normal  to  a  curve  C  at  a  point  Pi  on  C  is  the  line  through 
Pi  perpendicular  to  the  tangent  to  C  at  Pi.  Its  equation  is  found 
from  that  of  the  tangent  by  the  Rule  on  p.  114,  using  Theorem 

XII,  p.  117'. 


TANGENTS  AND  NORMALS  211 

Ex.  2.    Find  the  equation  of  the  normal  at  Pi  to  the  curve  in  Ex.  1. 

Solution.  The  equation  of  any  line  perpendicular  to  (2)  has  the  form 
(Theorem  XII,  p.  117) 

(3)  8  x  +  3  xth/  +  k  =  0. 

If  Pi  lies  on  this  line,  then  (Corollary,  p.  53) 
8a;i  +  Sxih/i-\-k  =  0, 
whence  A:  =  —  8  Xi  —  3  xi^yi. 

Substituting  in  (3),  the  equation  of  the  normal  is 

8  X  +  3  Xi2y  -  8  xi  -  3  iCi22/i  =  0. 

PROBLEMS 

1.  Eind  the  equations  of  the  tangents  and  normals  at  Pi(xi,  ?/i)  to  the 
curves  in  (a)  to  (e),  problem  2,  p.  209. 

Ans.   (a)  yiy  =  3  (x  +  Xi),  2/iX  +  3  y  =  Xiyi  +  3  ?/i. 

(b)  Xi^x  -  4  ?/ =  12  2/1,  4  X  +  Xi^y  =  4  xi  +  Xi^^i/i. 

(c)  xix  +  Viy  =  16,  yix  -  xiy  =  0. 

(d)  xix  -  yiy  =  4,  yix  +  x^y  =  2  x^yx. 

(e)  (3  Xi2  4-  2  xi)  X  -  2  2/1?/  -  Xi3  =  0,  2  yiX  +  (3  Xi2  +  2  Xi)  y  =  3  Xi^y i  +  4  Xi?/i. 

2.  Find  the  coordinates  of  a  point  on  each  of  the  curves  in  (/)  to  (j), 
problem  2,  p.  209,  and  then  find  the  equations  of  the  tangent  and  normal  at 
that  point. 

3.  Find  the  equations  of  the  tangents  and  normals  to  the  following  curves 
at  the  points  indicated. 

(a)  2/2-8x  +  4y=:0,  (0,  0).  Am.  2x  -  ?/ =  0,  x  +  2?/ =  0. 

(b)  xy  =  4,  (2,  2).  Ans.  x  +  y  =  4,  x  —  y  —  ^. 

(c)  x2-42/2  =  25,  Pi(xi,  ?/i).  Ans.  XiX-42/i?/=25,  42/iX+Xi2/=5xi?/i. 

(d)  x2  +  2xy  =  4,  Pi(xi,  ?/i). 

Ans.   (xi  +  2/i)  X  +  Xxy  -  4,  XiX  -  (xi  +  ^i)  y  =  x^  -  Xi?/i  -  y^. 

(e)  2/2  =  2 px.  Pi  (xi,  ?/i).    '      Ans.   2/i2/=P(a^  +  a:i),  2/iX+j32/  =  Xi?/i+i32/i. 

(f)  -0  +  ^  =  1'  Pi  (^1,2/1). 

^^^'   -^  +  -P"  -  ^'    62        a2  -     a262    '''^" 

(g)  62x2  -  a22/2  =  a262,  p^(xi,  yi). 

Ans.   62a; jic  _  a^y^y  =  a'^b^,  a^yiX  +  b^Xiy  =  (a2  +  62)  Xiyi. 
(h)  x^-y^  -\-x^  =  0,  (0,  0).  Ans.   y  =  ±x,  x  =  Ty- 


212 


ANALYTIC  GEOMETRY 


85.  Equations  of  tangents  and  normals  to  the  conic  sections. 
Theorem  I.    The  equation  of  the  tangent  to  the  circle 

C  :  x^  -\-  y^  =  r"^ 
at  the  point  P^  (x-^,  y^)  on  C  is 
(I)  oc^oo  +  ynj  =  r 


p2 


Proof.    Let  Pi  (xi,  ^i)  and  P2  (xi  +  A,  S'l  +  k)  be  two  points  on  the  circle  C. 
Then  (Corollary,  p.  63) 


(xi  +  W  +  (2/1  +  W  =  r% 


(1) 
and 

or 

(2)    Xi2  +  2xiA  +  /i2  4-yi2  +  22/iA;4-A;2  =  r2. 

Subtracting  (1)  from  (2),  we  have 
2  Xi/i  +  ^2  +  2  vik  +  A:2  =  0. 
Transposing  and  factoring,  this  becomes 
A;(2yi  +  fc)=-/i(2xi  +  h), 
k  _      2xi  4-  ^ 


whence 


22/i  +  A; 

is  the  slope  of  the  secant  through  Pi  and  P2. 

Letting  Pj.  approach  Pi,  h  and  k  approach  zero,  so  that  m,  the  slope  of  the 

tangent  at  Pi,  is 

,.    .,     -     2xi  +  A  xi 

m  =  limit  of = = -. 

2  2/1  +  A;  2/1 

The  equation  of  the  tangent  at  Pi  is  then  (Theorem  V,  p.  95) 

Xi 


y  -yi  = 


or 


(X  -  Xi  , 

2/1 
r2. 


Q.E.D. 


xix  +  yiy 
But  by  (1),  xi2  4-  yi2 

so  that  the  required  equation  is 

Xix  +  yiy 

Theorem  II.     The  equation  of  the  tangent  to  the  locus  of 
Ax^  +  Bxg  +  Ci/  +  Dx  -{-  Ey  +  F=0 
at  the  point  Pi  (xi,  3/1)  on  the  locus  is 

(II) 


TANGENTS  AND  NORMALS  213 

Proof.    Let  Pi  (xi,  2/1)  and  P2  (xi  +  ^,  ?/i  +  A:)  be  two  points  on  the  conic. 
Then  (Corollary,  p.  53) 

(3)  •      Axx^  +  Bxiv^  +  Cyi^  -\-  Dxi  +  Eyi  +  F  =  0  and 

A{xi  +  h)^  -{■B{xi  +  h)  {yi  +  A;)  -f  C (yi  +  k)^  +  D(xi  +  /i)  +  -E;(2/i  +  A;)  +  P=  0. 

Clearing  parentheses,  we  have 

(4)  Axi^  +  2  ^XiA  +  Ah'^  +  ^Xi2/i  +  -BxiA;  +  Byih  +  ^/iA; 

+  Cyi^  +  2  C2/1A;  -f  Ck^  4-  Dxi  +  -D^  +  ^2/1  +  Ek  +  F  =  0. 

Subtracting  (3)  from  (4),  we  obtain 

(5)  2Axih  +  Ah^  +  Bxik  +  5?/i/i  +  Bhk  +  2  CyiA;  -f-  Ck"^  -\-  Dh -\-  Ek  =  0. 

Transposing  all  the  terms  containing  h  and  factoring,  (5)  becomes 
k{Bxi  +  2Cyi-\-Ck  +  E)=-h{2Axi  -\- Ah  +  Byi  +  Bk-+  D), 
k  2Axi  + Byi  + D-{-Ah  + Bk 


whence 


h  Bxi  +  2  Cyi  +  E  -\-  Ck 


This  is  the  slope  of  the  secant  P1P2  [(1),  p.  207]. 

Letting  P2  approach  Pi,  h  and  k  will  approach  zero  and  the  slope  of  the 

^^^S^^t  i«  ^  ^  _  2Ax,  +  By,-i-D  ^ 

^  Bxi  +  2  C2ji -{- e' 

The  equation  of  the  tangent  line  is  then  (Theorem  V,  p.  95) 

2Axi-{-Byi  +  D.  , 

^'  Bxi  +  2Cyi  +  E^  '^ 

To  reduce  this  equation  to  the  required  form  we  first  cleaij  of  fractions  and 
transpose.     This  gives 

(2  Axi  +  By  I  +  D)x  fiBxi  +  2Cyi-\-E)y 

^-  (2  Axi^  +  2  Bxiyi  +  2  Cyi^  +  Dxi  +  ^2/i )  =  0. 

But  from  (3)  the  last  parenthesis  in  this  equation  equals 
-{Dxi  +  Eyi  +  2F). 

Substituting,  the  equation  of  the  tangent  line  is 
{2Axi  +  Byi  +-:©y:rH-  {Bxi  +  2Cyi  +  E)y  +  {Dxi  +  Eyi  +  2F)  =  0. 

Removing  the  parentheses,  collecting  the  coefficients  of  A,  B,  O,  D,  E^ 
and  P,  and  dividing  by  (2),  we  obtain  (II).  q.e.d. 

Theorem  II  enables  us  to  write  down  the  equation  of  the  taiP" 
gent  to  the  locus  of  any  equation  of  the  second  degree.  It  is 
remembered  most  easily  in  the  form  of  the  following  Rule. 


214  ANALYTIC  GEOMETRY 

Rule  to  write  the  equation  of  the  tangent  at  Pi  (xi,  t/i)  to  the  locus 
of  an  equation  of  the  second  degree. 

First  step.    Substitute  x^x  and  y^y  for  x"^  and  y"^,  — — - — —  for 

xy,  ana  — - — -  and  for  x  and  y  in  the  given  equation. 

Second  step.  Substitute  the  numerical  values  of  Xt  and  y^,  if  given, 
in  the  result  of  the  first  step.     The  result  is  the  required  equation. 

In  like  manner,  or  at  once  from  this  Kule,  we  have 
Theorem  III.     The  equation  of  the  tangejit  at  P^  (xi,  y^)  to  the 

ellipse  b^x^  +  a^y^  =  a%^   is  b^oc^oc  +  a^y^y  =  a^b^ ; 

hyperbola     b^x^  —  a^y"^  —  a%'^   is  b'^oc^uc  —  a^yiy  =  a^b"^ ; 
parabola  y"^  =  2px  is  y^y  =  _p  (x  +  x-^. 

By  the  method  on  p.  210,  we  obtain 

Theorem  IV.    The  equation  of  the  normal  at  Pi(xi,  y{)  to  the 
ellipse  b^x^  +  a^y^  =  a%^  is  ahj^oc  —  h^x^y  =  {a^  —  b^)  oc^yi ; 

hyperbola    b'^x^  —  a^y^  =  a%^  is  ahj^x  +  b'^x^y  —  {a^  +  6^)  x^y^ ; 
parabola  y"^  =  2px  is  y^x  +  j^y  =  x^y-i  +  vy^- 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  and  normals  to  the  following  conies 
at  the  points  indicated. 

(a)  3x2-10?/2  =  17,  (3,  1).  (d)  2x2 -^2 '=14,  (g^  _  2). 

(b)  y2  =  4aj,  (9,  _  6).  (e)  x^^hy'^  =  14,  (3,  1). 

(c)  x2  +  y2  ^  25,  (-  3,  -  4).  (f)  x2  =  62/,  (-  6,  6). 

(g)  x2-x2/  +  2x-7  =  0,  (3,2). 

(h)  X2/-y2  +  6x  +  82/-6  =  0,  (-1,4). 

The  directed  lengths  on  the  tangent  and  normal  from  the  point 
of  contact  to  the  Z-axis  are  called  the  length  of  the  tangent  and  the 
length  of  the  normal  respectively.  Their  projections  on  the  Z-axis 
are  known  as  the  subtangent  and  subnormal. 

2.  Find  the  subtangents  and  subnormals  in  (a),  (b),  (d),  and  (e),  prob- 
lem 1.  Am.    (a)   -  Lo,  _«_ ;  (b)  -  18,  2 ;  (d)  -  |,  6 ;  (e)  |,  -  f . 

3.  Find  the  lengths  of  the  tangents  and  normals  in  (a),  (b),Jd),  and  (e), 
probl&m  1.  Ans.   (a)  \  VTsT,  J^  VlsT ;  (b)  6  VTo,  2  VlO ; 

^  (d)  I  Vio,  2  VlO ;  (e)  \  V34,  |  Vsi". 

\ 


TANGENTS  AND  NORMALS 


216 


4.  Find  the  subtangents  and  subnormals  of  (a)  the  ellipse,  (b)  the  hyper- 
bola, (c)  the  parabola. 


Ans.   (a) 


Xi- 


62  a2  _  a;  2 

xi;  (b)^ , 

a^  Xi 


2xi,p. 


xi  a^  xi        a^ 

5.  Show  how  to  draw  the  tangent  to  a  parabola  by  means  of  the  sub- 
normal or  subtangent.  . 

6.  Prove  that  a  point  Pi  on  a  parabola  and  the  intersections  of  the 
tangent  and  normal  to  the  parabola  at  Pi  with  the  axis  are  equally  distant 
from  the  focus. 

7.  Show  how  to  draw  a  tangent  to  a  parabola  by  means  of  problem  6. 

8.  The  normal  to  a  circle  passes  through  the  center. 

9.  If  the  normal  to  an  ellipse  passes  through  the  center,  the  ellipse  is  a 
circle. 

10.  The  distance  from  a  tangent  to  a  parabola  to  the  focus  is  half  the 
length  of  the  normal  drawn  at  the  point  of  contact. 

11.  Find  the  equation  of  the  tangent  at  a  vertex  to  (a)  the  parabola; 
(b)  the  ellipse;  (c)  the  hyperbola. 

12.  Find  the  subnormal  of  a  point  Pi  on  an  equilateral  hyperbola. 

Ans.   Xi. 

13.  In  an  equilateral  hyperbola  the  length  of  the  normal  at  Pi  is  equal  to 
the  distance  from  the  origin  to  Pi. 

86.  Tangents  to  a  curve  from  a  point  not  on  the  curve. 

Ex.  1.    Find  the  equations  of  the  tangents  to  the  parabola  y"^  =  4x  which 
pass  through  P2  (—  3,  —  2). 

Solution.  Let  the  ;point  of 
contact  of  a  line  drawn  through 
P2  tangent  to  the  parabola  be 
Pi.  Then  by  Theorem  III  the 
equation  of  that  line  is 

(1)  ?/i?/ =  2  X  +  2  Xi. 

Since  P2  lies  on  this  line 
(Corollary,  p.  53), 

(2)  -2yi=-6  +  2xi; 
and  since  Pi  lies  on  the  parabola. 


(3) 


?/l2  =  4Xi. 


The  coordinates  of  Pi,  the 
point  of  contact,  must  satisfy 
(2)  and  (3).  Solving  them,  we 
find  that  Pi  may  be  either  of  the  points  (1,  2)  or  (9,  -  6). 


216  ANALYTIC  GEOMETKY 

If  (1,  2)  be  the  point  of  contact,  the  tangent  line  is,  from  (1), 

22/  =  2x  +  2, 
or  x-y  +  1  =  0. 

If  (9,  —  6)  be  the  point  of  contact,  the  tangent  line  is 
-  6  2/  =  2  X  +  18, 
or  x-Sy  +  9  =  0. 

The  method  employed  may  be  stated  thus : 

Rule  to  determine  the  equations  of  the  tangents  to  a  curve  C 
passing  through  P^ix^,  y^  not  on  C. 

First  step.  Let  P^  (x'l,  ?/i)  he  the  point  of  tangency  of  one  of  the 
tangents,  and  find  the  equation  of  the  tangent  to  C  at  P^  {Rule, 
p.  210). 

Second  step.  Write  the  conditions  that  (x^,  y^)  satisfy  the  result 
of  the  first  step  and  (cci,  y-^  the  equation  of  C,  and  solve  these  equa- 
tions for  Xi  and  yi. 

Third  step.  Substitute  each  pair  of  values  obtained  in  the  second 
step  in  the  result  of  the  first  step.  The  resulting  equations  are  the 
required  equations. 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  to  the  following  curves  which  pass 
through  the  point  indicated  and  construct  the  figure. 

(a)  x^  +  y^  =  25,  {7,  -1).  Ans.  3x  -  4?/ =  25,  4x  +  3?/ =  25. 

(b)  2/2  :=  4x,  (-  1,  0).  Ans.  y  =  x-\-\,y  +  x-\-l  =  0. 

(c)  16x2  +  25y2  =  400,  (3,  -  4).  Ans.  y  +  4  =  0,  Sx  -  2y  =  11. 

(d)  8y  =  x3,  (2,  0).  Ans.  y  =  0,27 x-8y- 54  =  0. 

(e)  x2  _}.  i6?/2  _  100  =  0,  (1,  2).  Ans.  None. 

(f)  2xy  +  ?/2  =  8,  (-  8,  8).        Ans.  2x  +  3?/  -  8  =  0,  4x  +  3?/  +  8  =  0. 

(g)  y^-\-4x-ey  =  0,  {-^,  -1).  Ans.  2x-Sy  =  0,  2x-y-\-2  =  0. 
(h)  x2  +  4  y  =  0,  (0,  -  6).  Ans.  None. 

(i)  x2  -  3 ?/2  +  2 X  +  19  =  0,  (- 1,  2). 

Ans.   x  +  3?/-5  =  0,  x-3y  +  7  =  0. 
(j)  y2  =  x^,  (f,  0).  Ans.   y  =  0,  3x  -  y  -  4  =  0,  3x  +  y  -  4  =  0. 

2.  Find  the  equations  of  the  lines  joining  the  points  of  contact  of  the 
tangents  in  (a),  (b),  (c),  (f),  (g),  and  (i),  problem  1. 

Ans.   (a)7x-y  =  25;    (b)  x  =  1 ;    (c)  12x  -  25y  =  100; 
(f)  x  =  l;    (g)  x-2y  =  0;   (i)  y  =  6. 


TANGENTS  AND  NORMALS 


217 


87.  Properties  of  tangents  and  normals  to  conies. 

Theorem  Y.    If  a  point  moves  off  to  infinity  on  the  parabola  y^ 
tangent  at  that  point  approaches  parallelism 
with  the  X-axis. 

Proof.    The  equation  of  the  tangent  at  the 
point  Pi(xi,  yi)  is  (Theorem  III,  p.  214) 
2/12/  =px+  pxi. 

Its  slope  is  (Corollary  I,  p.  86) 


2px,  the 


P 
m  =  —• 

2/1 
As  Pi  recedes  to  infinity  y^  becomes  infinite, 
and  hence  m  approaches  zero,  that  is,  the  tangent 
approaches  parallelism  with  the  X-axis.    q.  e  .  d. 

Theorem  VI.    If  a  point  moves  off  to  infinity  on  the  hyperbola 
b'^^-a'^y'^  =  aW, 
the  tangent  at  that  point  approaches  coincidence  with  an  asymptote. 

Proof.    The  equation  of  the  tangent  at  the  point  Pi  (Xi,  yi)  is  (Theorem  "* 
III,  p.  214) 
(1)  62a;ix  -  a'^yiy  =  a'^bK 

&2xi 


Its  slope  is  (Corollary  I,  p.  86) 


cC'yi 


As  Pi  recedes  to  infinity  Xi  and  yi  become  infinite  and  m  has  the  inde 
terminate  form  — . 

CO 

But  since  Pi  lies  on  the  hyperbola, 

62xi2  -  a2yi2  =  a'^b'^ 
Dividing  by  a'^yi^,  transposing,  and  extracting  the  square  root, 

6 


Multiplying  by 


bxi 
ayi 

62Xi 

'1 


2/1^ 


+  1. 


m  =  — —  =  ±-A  — +  1. 
a^yi  a  \  2/i2 


218 


ANALYTIC  GEOMETRY 


From  this  form  of  m  we  see  that  as  2/1  becomes  infinite  m  approaches 

±  -,  the  slopes  of  the  asymptotes  [(5),  p.  190],  as  a  limit.    The  intercepts  of 

(1)  are  —  and ■-     As  their  limits  are  zero  the  limiting  position  of  the 

tangent  will  pass  through  the  origin.     Hence  the  tangent  at  Pi  approaches 
coincidence  with  an  asymptote.  q.e.d. 

These  theorems  show  an  essential  distinction  between  the  form  of  the 
parabola  and  that  of  the  right-hand  branch  of  the  hyperbola. 

^  Theorem  vn.     The  tangent  and  normal  to  an  ellipse  bisect  respectively  the 
external  and  internal  angles  formed  by  the  focal  radii  of  the  point  of  contact  * 
Proof.    The  equation  of  the  lines  joining  Pi  (xi,  yi)  on  the  ellipse 

62x2  +  a22/2  =  a262 

to  the  focus  F'  (c,  0)  (Theorem  V,  p.  185)  is 
(Theorem  VII,  p.  97) 

yix  +  {c-xi)y-cyi  =  0, 
and  the  equation  of  PiP  is 
^  yix  -  {c  +  Xi)y  +  cT/i  =  0. 

The    equation    of    the    tangent    AB  is 
(Theorem  III,  p.  214) 

62xix  +  a^yiy  =  a^b^. 

We  shall  show  that  the  angle  6  which  AB  makes  with  PiF'  equals  the 
angle  0  which  PiP  makes  with  AB. 
By  Theorem  X,  p.  109, 

a2y^2  _  52cxi  -f  b'^xi^    _  {a'^yi^  +  b^Xr^  -  b^cxi 
b^xiyi  +  a^cyi  -  a^x^yi      a^cyi  -  {a^  -  62)  xi^i 

But  since  Pi  lies  on  the  ellipse. 


a2y^2  ^  52a;^2  _  (j252^ 


and  (Theorem  V,  p.  185) 

a262  -  62cxi 


Hence  tan  6  = 

In  like  manner 
tan  0  = 


a2-62 
62(a2- 


:C2. 
CXi) 


cyi 


< 


a'^cyi  -  c2xi2/i      cyi  {a^  -  cxi) 

-  62cxi  -  62xi2  -  a^i^      (62xi2  +  a^yr^)  +  62cxi 


h'^xiyi  -  a^cyi  -  a'^Xiyi 


a2c2/i  +  (a2-62)xi2/i 
aW  +  62cxi    _  62 
cyi 


a^cyi  +  c2xiyi 
*  This  theorem  finds  application  in  the  so-called  whispering  galleries. 


TANGENTS  AND  NORMALS 


219 


Hence  tan  ^  =  tan  0 ;  and  since  d  and  0  are  both  less  than  it,  6  =  <p. 
That  ia,  AB  bisects  the  external  angle  of  FPi  and  F'Pi,  and  hence,  also, 
CD  bisects  the  internal  angle.  q.b.d. 

In  like  manner  we  may  prove  the  following  theorems. 

*4llieorem  VIII.  The  tangent  and  normal  to  an  hyperbola  bisect  respec- 
tively the  internal  and  external  angles  formed  by  the  focal  radii  of  the  point 
of  contact. 


% 


Theorem  IX.  The  tangent  and  normal  to  a  parabola  bisect  respectively  the 
internal  and  external  angles  formed  by  the  focal  radius  of  the  point  of  contact 
and  the  line  through  that  point  parallel  to  the  axis* 

These  theorems  give  rules  for  constructing  the  tangent  and  normal  to  a 
conic  by  means  of  ruler  and  compasses. 

Construction.  To  construct  the  tangent  and  normal  to  an  ellipse  or  hyper- 
bola at  any  point,  join  that  point  to  the  foci  and  bisect  the  angles  formed  by 
these  lines.  To  construct  the  tangent  and  normal  to  a  parabola  at  any  point, 
draw  lines  through  it  to  the  focus  and  parallel  to  the  axis,  and  bisect  the 
angles  formed  by  these  lines. 

The  angle  which  one  curve  makes  with  a  second  is  the  angle  which  the 
tangent  to  the  first  makes  with  the  tangent  to  the  second  if  the  tangents  are 
drawn  at  a  point  of  intersection. 

N  Theorem  X.    Confocal  ellipses  and  hyperbolas  intersect  at  right  angles. 
Proof.    Let 

+  rr  =  1    and   — :  —  rrr  =  1 


(2) 


a2      62  a'2     6'2 

be  an  ellipse  and  hyperbola  with  the  same  foci.     Then 

(3)  a2  -  62  =  a/2  +  5^2. 

For  if  the  foci  are  (±  c,  0),  then  in  the  ellipse  c^  =  a^-  b^  and  in  the  hyperbola  C'=a'^  +  b'' 
(Theorems  V  and  VI,  p.  185). 


*  This  theorem  finds  application  in  reflectors  for  lights. 


220  ANALYTIC  GEOMETRY 

The  equations  of  the  tangents  to  (2)  at  a  point  of  intersection  Pi  {xi,  y\) 
are  (Rule,  p.  214) 

It  is  to  be  proved  that  the  lines  (4)  are  perpendicular,  that  is  (Corollary 
III,  p.  87),  that 


(5)  -^-J^  =  0. 

^  ^  a2a'2      hW^ 

Since  P\  lies  on  both  curves  (2),  w^e  have 

^V^'  =  land^-^  =  1. 

a2  &2  (j'2         5'2 

Subtracting  these  equations,  we  obtain 
,g.  (a^-a^-^)xi2  _  (62  +  y2)y^2  ^  ^ 

But  from  (3),  a'^  -  a'2  =  lfi  ^  h'\ 

and  hence  (6)  reduces  to  (5)  and  the  lines  (4)  are  perpendicular.  q.e.d. 

In  like  manner  we  prove 

Nrheorem  XI.    Two  parabolas  with  the  same  focus  and  axis  which  are  turned 
in  opposite  directions  intersect  at  right  angles. 

Hence  the  confocal  systems  in  section  82,  p.  200  (Ex.  4  and  problem  6), 
are  such  that  the  two  curves  of  the  system  through  any  point  intersect  at 
right  angles. 

PROBLEMS 

1.  Tangents  to  an  ellipse  and  its  auxiliary  circle  (p.  206)  at  points  with 
the  same  abscissa  intersect  on  the  X-axis. 

2.  The  point  of  contact  of  a  tangent  to  an  hyperbola  is  midway  between 
the  points  in  which  the  tangent  meets  the  asymptotes. 

3.  The  foot  of  the  perpendicular  "from  the  focus  of  a  parabola  to  a  tan- 
gent lies  on  the  tangent  at  the  vertex. 

4.  The  foot  of  the  perpendicular  from  a  focus  of  a  central  conic  to  a 
tangent  lies  on  the  auxiliary  circle  (p.  206). 

6 .  Tangents  to  a  parabola  from  a  point  on  the  directrix  are  perpendicular 
to  each  other. 

6.  Tangents  to  a  parabola  at  the  extremities  of  a  chord  which  pass 
through  the  focus  are  perpendicular  to  each  other. 

7.  The  ordinate  of  the  point  of  intersection  of  the  directrix  of  a  parabola 
and  the  line  through  the  focus  perpendicular  to  a  tangent  is  the  same  as  that 
of  the  point  of  contact. 


TANGENTS  AND  NORMALS 


221 


8.  How  may  problem  7  be  used  to  draw  a  tangent  to  a  parabola  ? 

9.  The  line  drawn  perpendicular  to  a  tangent  to  a  central  conic  from  a 
focus  and  the  line  passing  through  the  center  and  the  point  of  contact  inter- 
sect on  the  corresponding  directrix. 

10.  The  angle  which  one  tangent  to  a  parabola  makes  with  a  second  is 
half  the  angle  which  the  focal  radius  drawn  to  the  point  of  contact  of  the 
first  makes  with  that  drawn  to  the  point  of  contact  of  the  second. 

11.  The  product  of  the  distances  from  a  tangent  to  a  central  conic  to  the 
foci  is  constant. 

12.  Tangents  to  any  conic  at  the  ends  of  the  latus  rectum  (double  chord 
through  the  focus  perpendicular  to  the  principal  axis)  pass  through  the 
intersection  of  the  directrix  and  principal  axis. 

13.  Tangents  to  a  parabola  at  the  extremities  of  the  latus  rectum  are 
perpendicular. 

14.  The  equation  of  the  parabola  referred  to  the  tangents  in  problem  13  is 

x2  -  2a;?/  +  2/2  -  2  V2p {x  +  y)  +  2p2  =  o, 

or  (compare  p.  17)  x^  -\- y^  =  yp  V2. 

15.  The  area  of  the  triangle  formed  by  a  tangent  to  an  hyperbola  and 
the  asymptotes  is  constant. 

16.  The  area  of  the  parallelogram  formed  by  the  asymptotes  of  an 
hyperbola  and  lines  drawn  through  a  point  on  the  hyperbola  parallel  to 
the  asymptotes  is  constant. 

88.  Tangent  to  a  curve  at  the  origin.    If  a  curve  passes  through 
the  origin,  the  equation  of  the  tangent  at 
that  point  is  easily  found. 

Ex.  1.  Find  the  equation  of  the  tangent  at  the 
origin  to 

Solution.  To  find  the  slope  of  the  tangent  at 
Pi(0,  0),  let  P2(0  +  A,  0  +  k)  be  a  second  point 
on  C.  The  conditions  that  Pi  and  P2  lie  on  C 
give  but  one  equation, 

whence  the  slope  of  the  secant  P1P2  is  [(1),  p.  207] 


m=-=-2 
h 


^2. 


Letting  P^  approach  Pi,  h  and  k  approach  zero, 
and  the  slope  of  the  tangent  is  the  limit  of  m, 
which  is  —  2. 


222  ANALYTIC  GEOMETRY 

Hence  the  equation  of  the  tangent  is  (Theorem  I,  p.  58) 

or  2  X  +  y  =  0. 

Notice  that  this  equation  may  be  obtained  at  once  by  setting  the  terms  of 
the  first  degree  in  the  equation  of  C  equal  to  zero. 

If  a  curve  passes  through  the  origin,  the  constant  term  in  its 
equation  must  be  zero  (Theorem  YI,  p.  73),  so  that  its  equation 
must  have  the  form 

Ax  +  By  +  Cx^  +  Dxi/  -\-  Eif  +  Fx^ -] =  0, 

where  the  dots  indicate  that  there  may  be  other  terms  whose 
degree  in  x  and  y  may  be  three  or  greater. 

Theorem  XII.  The  equation  of  the  tangent  at  the  origin  to  the 
curve  C  whose  equation  arranged  according  to  ascending  powers  of 
X  and  y  is 

Ax  +  By-{-  Cx'^  +  Dxy  -{■  Ey'^  +  Fx^  ^ =  0, 

is  Ax  -{-  By  =  0. 

That  is,  the  equation  of  the  tangent  to  C  at  the  origin  is  obtained 

by  setting  equal  to  zero  the  terms  of  the  first  degree  in  x  and  y. 

Proof  Pi(0,  0)  lies  on  C.  Let  Pi{h,  k)  be  a  second  point 
on  C.     Then  (Corollary,  p.  53) 

Ah  -^Bk-\-  Ch^  +  Dhk  +  Ek'^  +  F/^s  +  . . .  =  Q. 
Transposing  all  terms  containing  h,  and  factoring, 

k{B  +  Ek-\ )  =  -h{A  +  Ch  +  Dk  +  Fh''  +  •••). 

k  _      A-\-Ch^-  Dk-\-  Fh''  +  > .  • 
"  h~  B-\-  Ek-\ 

Letting  P^  approach  P^,  the  limit  of  —^  which  is  the  slope  of 

A 

the  tangent,  is  seen  to  be 

Hence  the  equation  of  the  tangent  is  (Theorem  V,  p.  95) 

A 

y  = X,  — 

^  B    ^ 

or  .  Ax  +  By  =  0.  q.e.d. 

If  ^  =  0  and  B  =  0,  the  terms  of  the  lowest  degree,  if  set  equal  to  zero,  will 
be  the  equation  of  the  two  or  more  lines  which  will  then  be  tangent  to  C  at 
the  origin.  For  example,  if  the  equation  of  C  is  x'^  —  y'^  +  x^  —  0,  the  two  lines 
a;2  _  y2  —  0  will  be  tangent  to  C  at  the  origin  (problem  3,  (h),  p.  211). 


TANGENTS  AND  NORMALS 


223 


89.  Second  method  of  finding  the  equation  of  a  tangent.   The 

tangent  to  a  curve  C  at  a  point  Pi  may  now  be  found  as  follows. 
Transform  C  by  moving  the  origin  to  Pi  (Theorem  I,  p.  160). 
The  equation  of  the  tangent  at  Pi  in  the  new  coordinates  is  then 
found  immediately  by  Theorem  XII.  Transform  it  by  translating 
the  axes  to  their  first  position.  The  result  is  the  equation  of  the 
tangent  at  Pi  in  the  given  coordinates. 

Ex.  1.    Find  the  equation  of  the  tangent  to  C  :  4  x^  —  2  ?/2  _j.  3^3  =  q  at 
Pi(-  2,  2)  which  lies  on  C. 

Solution.    Set  (Theorem  I,  p.  160) 

x  =  x'-2,  y  =  y'  +  2. 
The  equation  of  C  becomes 
4  (x'  -  2)2  -  2  (y'  +  2)2  +  {x'  -  2)3  =  0. 
Only  the   terms  of  the  first  degree  are 
needed,  and  these  may  he  picked  out  without 
clearing  the  parentheses.     Tlie  equation  of 
the  tangent  is  therefore 

4  (-  4 x')  -  2  . 4  ?/'  +  12  x'  =  0, 
or  x'  +  2  y'  —  0. 

To  transform  to  the  old  axes,  set 
x'  =  X  +  2,  y'  =  y  -2. 
"We  thus  obtain 

x  +  2?/-2  =  0, 
which  is  the  equation  of  the  tangent  to  C  at  Pi. 


PROBLEMS 

1.  Find  the  equations  of  the  tangents  at  the  origin  to 

(a)  x2  +  2  xy  +  ?/2  -  6  X  +  8  y  =  0.  (d)  y  =  x^  -  2  x2  +  x. 

(h)  xy  -y^  -\-x-Sy  =  0.  {e)  x^  -\- y^  +  x  -  y  =  0. 

(c)  x2  +  4x2/  -  3x  +  4?/  =  0.  (f)  x3  +  x2  -  Sxy  -  4y2  =  q. 

2.  Find  the  equations  of  the  tangents  to  the  following  curves  at  the  points 
indicated  by  the  method  of  section  89. 

(a)  9x2  -  2/2  +  2 X  -  4  =  0,  (2,  6). 

(b)  x2  +  4xy  +  62/-7=0,  (-1,  3). 

(c)  xy  +  6x  -  4?/  -  6  =  0,  (2,  3). 

(d)  y2a.4x  +  2y  +  8  =  0,  (-4,  2). 

(e)  y^  =  x^  +  8,  (2,  4). 

(f)  2/  =  X*  -  3x3  -  5x2  +  4x  +  4,  (0,  4). 


Ans.  19x-6y-2  =  0. 

Ans.  5x  +  2/  +  2=0. 

Ans.  9x-2y-12  =  0. 

Ans.  2x  +  3y  +  2  =  0. 

Ans.  Sx-2y  +  2  =  0. 

An^.  y  =  4  X  +  4. 


224  ANALYTIC  GEOMETRY 

3 .  Find  the  angle  which  the  locus  of  xy-\-4y  —  2x  =  0  makes  at  the 
origin  with  that  of  x^  -\-  ixy  +  x -{-  8y  =  0.  j^^^^    5 

■    4* 

4.  Find  the  angle  which  the  line  2x  —  Sy  —  9  =  0  makes  with  the  locus 
of  x2/  +  6x-4y-19  =  0at(3,  -1).  ^^^    ^ 

4 

MISCELLANEOUS   PROBLEMS 

1.  Find  the  equations  of  the  tangents  and  the  normals  to  the  following 
conies  at  the  points  indicated. 

(a)  x^  +  4xy  -ix-lOy  +  7  =  0,  (3,  -  2). 

(b)  x?/ -4x4-32/ -4  =  0,  (-1,4). 

(c)  x?/  +  2/2  +  2x  +  2?/  =  0,  (-  3,  3). 

(d)  2/2  +  4x  +  62/  -  27  =  0,  (5,  -  7). 

(e)  x2  +  3X2/  +  2/2  -  102/  -  1  =  0,  (2,  3). 

(f)  x2  -  8x  +  32/ -  14  =  0,  (1,  7). 

2.  Find  the  equation  of  one  of  the  tangents  to  the  ellipse  x^  +  9?/^  —  4x 
4-  9 2/  =  0  which  is  parallel  to  the  line  4x  —  9y  —  36  =  0. 

.3.  For  what  point  of  the  parabola  y'^  =  2px  is  the  length  of  the  tangent 
equal  to  four  times  the  abscissa  of  the  point  of  contact  ? 

4.  What  is  the  length  of  the  tangent  to  a  parabola  at  an  extremity  of 
the  latus  rectum  ?  Restate  the  equation  of  the  parabola  in  problem  14, 
p.  221,  in  terms  of  this  length. 

5.  For  what  point  on  the  parabola  y^  =  2px  is  the  normal  equal  to 
(a)  twice  the  subtangent?  (b)  the  difference  between  the  subtangent  and 
the  subnormal? 

6.  Through  a  point  of  the  ellipse  h^"^  +  a'^y^  =  a'^b'^  and  that  point  of 
the  auxiliary  circle  with  the  same  abscissa  normals  are  drawn.  What  is 
the  ratio  of  the  subnormals  ? 

7.  For  what  points  of  an  hyperbola  is  the  subtangent  equal  to  the 
subnormal  ? 

8.  The  ordinate  of  a  point  on  an  equilateral  hyperbola  and  the  length  of 
the  tangent  drawn  from  the  foot  of  that  ordinate  to  the  auxiliary  circle  are 
equal. 

9.  A  tangent  to  a  parabola  meets  the  directrix  and  latus  rectum  produced 
at  points  equally  distant  from  the  focus. 

10.  The  semi-conjugate  axis  of  a  central  conic  is  a  mean  proportional 
between  the  distance  from  the  center  to  a  tangent  and  the  length  of  the 
normal  drawn  at  the  point  of  contact. 


TANGENTS  AND  NORMALS  225 

11.  Find  the  points  of  the  ellipse  for  which  the  lengths  of  the  tangent 
and  normal  are  equal. 

12.  Any  point  on  an  equilateral  hyperbola  is  the  middle  point  of  that  part 
of  the  normal  included  between  the  axes  of  the  hyperbola. 

13.  A  circle  is  drawn  through  a  point  on  the  minor  axis  of  an  ellipse  and 
through  the  foci.  Show  that  the  lines  drawn  through  the  given  point  and 
the  points  of  intersection  of  the  circle  and  ellipse  are  normal  to  the  ellipse. 

14.  How  many  normals  may  be  drawn  through  a  given  point  to  (a)  an 
ellipse  ?  (b)  an  hyperbola  ?  (c)  a  parabola  ? 


CHAPTER    X 

RELATIONS  BETWEEN  A  LINE  AND  A  CONIC.    APPLICA- 
TIONS OF  THE   THEORY  OF  QUADRATICS 

90.  Relative  positions  of  a  line  and  conic.  If  a  line  and  conic 
are  given,  it  is  evident  that 

(a)  the  line  is  a  secant  of  the  conic, 

(b)  the  line  is  tangent  to  the  conic,  or 

(c)  the  line  does  not  meet  the  conic. 

The  coordinates  of  the  points  of  intersection  of  the  line  and 
conic  are  found  by  solving  their  equations  (Rule,  p.  76),  which 
are  of  the  first  and  second  degrees  respectively.     To  solve,  we 
eliminate  y*  and  arrange  the  resulting  equation  in  the  form 
(1)  •'  Ax^  +  Bx  +  C  =  0. 

Denote  the  roots  by  Xi  and  X2  and  the  discriminant  B^—AAChj  A. 
Analytically  the  three  cases  above  present  themselves  as  follows  : 

(a)  If  A  is  positive,  the  line  is  a  secant. 

For  Xi  and  X2  are  real  and  unequal  (Theorem  II,  p.  3),  and  hence  they  are  the 
abscissas  of  the  points  of  intersection,  which  must  be  distinct. 

(h)  If  A  is  zero,  the  line  is  a  tangent. 

For  in  this  case  a^i  =  x^,  so  that  the  points  of  intersection  coincide. 

(c)   If  A  is  negative,  the  line  does  not  meet  the  conic. 

For  Xx  and  Xg  are  imaginary,  and  hence  there  are  no  points  of  intersection 
(p.  77). 

If  ^  =  0,  one  root  of  (1)  is  infinite  (Theorem  IV,  p.  15)  and  one  point  of  inter- 
section is  said  to  be  "  at  infinity." 

li  A  =  0  and  B  =  0,  then  both  roots  of  (1)  are  infinite  and  the  line  is  said  to  be 
**  tangent  at  infinity." 

If  ^  =  0,  B  =  0,  and  C  =  0,  then  (1)  is  satisfied  by  all  values  of  x,  and  hence 
has  an  infinite  number  of  roots.  All  of  the  points  on  the  line  lie  on  the  conic; 
that  is,  the  conic  is  degenerate  and  consists  of  straight  lines  of  which  the  given 
line  is  one. 

*If  one  equation  does  not  contain  ?/,  then  x  is  found  by  solving  that  equation.  But 
for  our  purposes  it  is  unnecessary  to  complete  the  solution. 

226 


LINE  AND  CONIC 


22T 


"*»^         ^ 

r/. 

/ 

^"v 

^       7 

Ns7 

%\ 

Z      ^ 

X' 

J    %% 

_J 

^  y 

7 

y 

2-^ 

^ 

^^ 

-7^ 

—  y'— 

In  solving  the  equations  of  the  line  and  conic  it  might  be  easier 

to  eliminate  x  than  y.     Then  (1)  would  be  a  quadratic  in  y,  but 

the  result  of  the  discussion  would  be  the  same. 

If  one  equation  did  not  contain  y,  it  would  be  necessary  for  our  purposes  to 
eliminate  x  instead  of  y,  and  vice  versa. 

Ex.  1.  Determine  the  relative  positions  of  the 
line  3x  —  2y  +  6  =  0  and  the  parabola  y^  +  4x  =  0. 

Solution.    It  is  easier  to  eliminate  x  than  y. 

Solving  the  equation  of  the  line  for  x,  we  obtain 

2y  -Q 

x=  — -. 

3 

Substituting  in  y2  +  4  x  =  0,  we  get 

3  2/2  +  8  y  -  24  =  0. 
The  discriminant  of  this  quadratic  is 
A  =  82-4..8(-24)  =  352. 
As  A  is  positive,  the  line  is  a  secant. 

Ex.  2.  Determine  the  relative  position  of  the  line  4  x  +  y 
+  5  =  0  and  the  ellipse  9  ic^  +  ?/2  =  9. 

Solution.  It  is  easier  to  eliminate  y  than  x.  From  the 
first  equation, 

y  =-  (ix+  5). 

Substituting  in  the  second  and  arranging,  we  get 
25x2 +  40X  + 16  =  0. 

The  discriminant  is  A  =  402  -  4  •  25  •  16  =  0.  Hence  the 
line  is  a  tangent. 

Ex.  3.  Determine  the  relative  position  of  the  loci  ofx2  —  2/2^3a;  —  3y  =  0 
and  X  —  y  =  0. 

Solution.    Eliminating  y,  we  get 

x2  -  x2  4-  3  X  -  3  X  =  0, 

or  0  •  x2  +  0  .  X  +  0  =  0. 

As  this  equation  is  true  for  all  values  of  x, 
then  all  of  the  points  on  the  line  lie  on  the  conic. 
The  equation  of  the  conic  may  evidently  be 
written  {x  —  y)  (x  +  y  +  S)  =  0.  The  locus  of 
this  equation  is  (Theorem,  p.  66)  the  degenerate 
conic  consisting  of  the  pair  of  lines 

x-y  =  0,  x  +  y  +  S  =  Or 
of  which  one  is  the  given  line. 


..[    U:4 

\ 

^ 

A'  Mo 

X 

J 

y] 

[ 

Ans. 

Tangent. 

Ans. 

Do  not  meet. 

Ans. 

Secant. 

Ans. 

Tangent. 

Ans. 

Tangent. 

Ans. 

Secant. 

Ans. 

Do  not  meet. 

Ajis. 

Do  not  meet. 

Ans. 

Secant. 

Ans. 

Tangent. 

228  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Determine  the  relative  positions  of  the  loci  of  the  following  equations 
and  plot  their  loci. 

(a)  x  +  y  +  l  =  0,x^  =  4y. 

(b)  x-2y  +  20  =  0^x^  +  y^  =  16. 

(c)  y2-4x  =  0,2x  +  3y-S  =  0. 

(d)  x2  +  2/2  -  X  -  2  ?/  =  0,  X  +  2  2/  =  5. 

(e)  2xy  -Sx  -y  =  0,y  +  Sx-6  =  0. 

(f )  x^  +  y^  -  Gx  -  Sy  =  0,  X  -  2y  =  6. 

(g)  4  x2  +  ^-2  _  16  aj  =  0,  X  +  ?/  -  8  =  0. 
(h)  x2  +  ?/2  -  8x  -  6  =  0,  X  +  8  =  0. 

(i)  8x2-6  2/2  +  i6x-32  =  0,  2x-3y  =  0. 

(j)  x2  +  xy  +  2x  +  y  =  0,  2x  +  y  +  4  =  0. 
(k)  x2  +  2  x?/  +  ?/2  +  4  X  -  4  ?/  =  0,  X  +  ?/  =  1. 

Ans.    Secant,  with  one  point  of  intersection  at  infinity. 

(1)  4 x2  —  ?/2  +  4  X  +  1  =  0,  2  X  —  2/  +  1  =  0,      Ans.    Line  is  part  of  conic, 
(m)  x2  +  4x!/  +  ?/2  +  4 X  -  6 ?/  =  0,  2 X  -  3 ?/  =  0.  Ans.    Tangent, 

(n)  x2  -  4  ?/2  +  8  ?/  -  20  =  0,  X  -  2  ?/  +  2  =  0.     Ans.    Tangent  at  infinity, 
(o)  x2-6x?/  +  0?/2  +  x-3?/-2  =  0,  x-32/=l. 

Ans.    Line  is  part  of  conic, 
(p)  6  x2  —  5  x?/  —  6  ?/2  =  18,  2  X  —  3  ?/  =  0.  Ans.    Tangent  at  infinity. 

2.  Find  the  middle  points  of  the  chords  of  the  conies  in  (c),  (f),  and  (i), 
problem  1,  which  are  formed  by  the  given  line. 

Ans.   (c)  (V-,  -3);    (f)  (-V-,  -  |) ;    (i)  (- f,  -1). 

3.  Interpret  Theorem  II,  p.  3,  geometrically  by  determining  the  relative 
positions  of  the  parabola  y  =  Ax^  +  Bx  +  C  and  the  line  y  =  0.  Construct 
the  figure  if 

(a)  A  =  l,  B=-l,  C=0;   (b)  A  =  l,  B=C=0;   (c)  A  =  l  B=l,  C=0. 

91.  Relative  positions  of  lines  of  a  system  and  a  conic,  and 
of  a  line  and  conies  of  a  system.  Given  a  system  of  lines  (that 
is,  an  equation  of  the  first  degree  containing  a  parameter  k)  and 
a  conic,  we  can  determine  the  values  of  k  for  which  the  lines  of 
the  system  intersect,  are  tangent  to,  or  do  not  meet  the  conic,  as 
follows. 

Eliminate  x  or  y,  as  may  be  more  convenient,  from  the  equa- 
tions of  the  system  of  lines  and  the  conic,  thus  obtaining  an 
equation  either  of  the  form 

(1)  Aif -\- By -\- C  =  0  01  Ax^  +  Bx +  0  =  0. 


LINE  AND  CONIC 


229 


The  discriminant  A  will  be  in  general  a  quadratic  in  k. 
Determine  the  values  of  k  for  which  A  is  positive,  zero,  or 
negative  (Theorem  III,  p.  11)  and  apply  the  results  of  the 
preceding  section. 

The  same  process  serves  to  separate  the  conies  of  a  system 
(that  is,  the  loci  of  an  equation  of  the  second  degree  containing 
a  parameter  k)  into  three  classes  according  as  they  intersect,  are 
tangent  to,  or  do  not  meet  a  given  line.  Only  here  the  values  of  A;, 
if  any,  for  which  the  equation  has  no  locus  must  be  excluded. 

Ex.  1.  Find  the  values  of  k  for  which  the  line  y  =  2x  -\-  k  intersects,  is 
tangent  to,  or  does  not  meet  the  ellipse  x2  +  4?/2  —  8x  +  4?/  =  0. 

Solution.    Eliminating  y  by  substitution  in  the  second  equation,  we  obtain 

17x2  +  16A;x  +  4fc24.4A:  =  0. 
The  discriminant  of  this  quadratic  is 

A  =  (16  A:)2  _  4  .  17  (4  A:2  +  4A;)  =-  16(fc2  +  17  k). 


17, 


By  (a),  (&),  and  (c),  p.  226, 

(a)  the  line  is  a  secant  if  -  16  {k^  +  17  fc)  >  0  ; 

(b)  the  line  is  a  tangent  if  -  16  {k^  +  17  A:)  =  0 ; 

(c)  the  line  does  not  meet  the  ellipse  if  —  16  (A;^  +  17  A;)  <  0. 

Apply  Theorem  III,  p.  11,  to  the  quadratic  —  16(fc2  +  17  A;). 

Since  A  =  (—  16  •  17)^  is  positive,  ^  =  —  16,  and  the  roots  are  0  and 

(a)  if  -  17  <  A;  <  0,  the  quadratic  -  16  (A:^  +  17  A;)  >  0  ; 

(b)  if  A:  =  0  or  -  17,  the  quadratic  -  16  {k^  +  17  A;)  =  0  ; 

(c)  if  A;  <  -  17  or  A;  >  0,  the  quadratic  -  16  (A;2  +  17  A:)  <  0. 

Hence 

(a)  the  line  is  a  secant  if  —  17  <  A:  <  0. 

(b)  the  line  is  a  tangent  if  A:  =  0  or  —  17. 

(c)  the  line  does  not  meet  the  ellipse  if  A:  <  —  17  or  A;  >  0. 

The  lines  of  the  system  are  all  parallel.    The  figure  shows  the  two  tangent 
lines  and  indicates  where  the  lines  lie  for  different  values  of  k. 


230  ANALYTIC  GEOMETRY 

PROBLEMS 

1 .  Determine  the  values  of  k  for  which  the  loci  of  the  following  equations 
(a)  intersect,  (b)  are  tangent,  (c)  do  not  meet.     Construct  the  figure  in  each 


(a)  2/  =  A:x  —  1,  x2  =  4  2/. 

Ans.   (a)  A;>1  or  ^<-l;   (b)  A;  =  ±  1 ;    (c)  -l<k<\. 

(b)  X  +  2  y  =  A;,  x2  +  2/2  =  5. 

Ans.   (a)  -  5  <  fc  <  5  ;    (b)  A:  =  ±  5  ;    (c)  A:  >  5  or  A;  <  -  5. 

(c)  x'^  +  y^  =  k,  3  X  -  4  2/  +  10  =  0. 

Ans.    (a)  A:  >  4  ;    (b)  A:  =  4  ;    (c)  0  <  A;  <  4. 

(d)  y  =  A;x  +  2,  x2  —  8  y  =  0.  Ans.    (a)  For  all  values  of  k. 

(e)  x2  +  y2  _  2  A;x  =  0,  y  =  x. 

Ans.    (a)  For  all  values  except  A;  =  0  ;   (b)  A;  =  0. 

(f)  4  x?  -  2/2  =  16,  y  =  A:x. 

Ans.   (a)  -  2  <  A;  <  2 ;    (b)  A:  =  ±  2  ;    (c)  A;  >  2  or  A:  <  -  2. 

(g)  y'^  =  2kx,  X  -  2  2/  +  2  =  0. 

Ans.   (a)  A;  >  1  or  A;  <  0 ;   (b)  A:  =  0  or  1;  (c)  0  <  A:  <  1. 
(h)  x2  +  4 ^2  _  8x  =  0,  y  =  kx  +  2-^k. 

Ans.  (a)  All  values  except  A:  =  0  ;  (b)  fc  =  0. 
(i)  xy  =  k,  2 X  +  y  +  4  =  0.  Ans.  (a)  A;  <  2  ;  (b)  Ar=:  2  ;  (c)  A;  >  2. 
(j)  X2/.  +  2/2  -  4  X  +  8  2/  =  0,  X  -  2  2/  +  A:  =  0. 

Ans.   (a)  A;  >  48  or  A;  <  0  ;    (b)  A;  =  0  or  48  ;    (c)  48  >  A;  >  0. 
(k)  4x2  +  2/2  -  6x  +  6y  =  0,  2/  =  A;x  +  1  -  A;. 

Ans.    (a)  A;  >  1  or  A;  <  -  If  ;    (b)  A:  =  1  or  -  if ;    (c)  -  if  <  A;  <  1. 

2.  Determine  the  values  of  k  for  which  the  loci  of  the  following  equations 
are  tangent  and  construct  the  figure. 

(a)  x2  -  4  2/  +  16  =  0,  2/  =  A;. 

(b)  9x2  +  162/2  =  144^  y  _  a-  ^  ^^ 

(c)  4x2/  +  y2  +  16  _  0,  X  =  A;. 

(d)  x2  +  4x2/  +  2/2  =  A:,  2/  =  2x  +  1. 

(e)  x2  +  2x2/  +  2/2  +  8x-62/  =  0,  4:X-^y  =  k. 

(f)  x2  +  2x2/ -4x  +  22/  =  0,  2x  -  2/ +  A;  -  3  =  0. 

92.  Tangents  to  a  conic.  If  in  the  preceding  section  the  value 
of  the  discriminant  of  (1)  is  zero,  then  the  line  and  conic  are  tan- 
gent. The  equation  obtained  by  setting  that  discriminant  equal 
to  zero  is  called  the  condition  for  tangency.  Hence  the  condition 
for  tangency  of  a  line  and  conic  is  found  by  eliminating  either 
X  or  y  from  their  equations  and  setting  the  discriminant  of  the 
resulting  quadratic  equal  to  zero. 

Thus  in  Ex.  1,  p.  229,  the  condition  for  tangency  is  A  =  -  16  (A;2  +  17  A;)  =  0. 


Ans. 

k  =  4. 

Ans. 

k=±5. 

Ans. 

k=±2. 

Ans. 

k=--i-,. 

Ans. 

k  =  0. 

An^. 

A;  =  3  or  13 

LINE  AND  CONIC 


231 


X      v 
Ex.  1.    Find  tlie  condition  for  tangency  of  the  line  -  +  -  =  1  and  the  parab- 
ola 2/2  =  2px. 

Solution.  Eliminating  z  by  solving  the  first  equation  for  x  and  substituting 
in  the  second,  we  get 

6^2  +  2  apy  —  2  abp  =  0. 

The  discriminant  of  this  quadratic  is 

A  =  (2  ap)2  -  4  6  (-  2  abp)  =  4  op  (ap  +  2  62). 
Hence  the  condition  for  tangency  is 

4 ap{ap  +  2 62)  =  0  or  ap {ap  +  2 62)  =  0. 

Rule  to  find  the  equation  of  a  line  tangent  to  a  given  conic  and 
satisfying  a  second  condition. 

First  step.  Write  the  equation  of  the  system  of  lines  satisfying 
the  second  condition. 

Second  step.  Find  the  condition  for  tangency  of  the  equation 
found  in  the  first  step  and  the  given  conic. 

Third  step.  Solve  the  equation  found  in  the  second  step  for  the 
value  of  the  parameter  of  the  system  of  lines  and  substitute  the  real 
values  found  in  the  equation  of  the  system.  The  equations  obtained 
are  the  required  equations. 

Ex.  2.  Find  the  equations  of  the  lines  with  the  slope  \  which  are  tangent 
to  the  hyperbola  x2  —  6  y2  +  12  y  —  18  =  0  and  find  the  points  of  tangency. 

Solution.    First  step.    The  lines  of  the  system 
(1)  y  =  ix  +  fc  . 

have  the  slope  \  (Theorem  I,  p.  58). 


y\ 

y 

V 

^ 

y^ 

^ 

^ 

^ 

X 

s, 

^ 

y 

^ 

X" 

(^ 

b 

\ 

^ 

^ 

^ 

t 

I 

^ 

t^ 

X 

\ 

\ 

X 

' 

X 

^ 

^ 

-f) 

s 

V 

fi 

"^ 

^ 

T-e 

■1 

^ 

X 

*• 

, 

i 

Second  step.    Solving  (1)  for  x  and  substituting  in  the  given  equation, 
(2)  y2  +  (4  ^  _  6)  y  +  9  -  2  A:2  =  0. 

Hence  the  condition  for  tangency  is 

(4  fc  -  6)2  -  4  (9  -  2  A:2)  =  0. 


232  ANALYTIC  GEOMETEY 

Third  step.    Solving  this  equation,  A;  =  0  or  2. 
Substituting  in  (1),  we  get  the  required  equations,  namely, 

(3)  x-2?/  =  0,  x-2?/  +  4  =  0. 

To  find  the  points  of  tangency  we  substitute  each  value  of  k  in  (2),  which 
then  assumes  the  second  form  of  (7),  p.  4,  namely, 

if  fc  =  0,  (2)  becomes  (?/  -  3)2  =  0  ;  .-.  y  =  Z; 
if  A:  =  2,  (2)  becomes  (y  +  1)2  =  0  ;.-.?/  =  -  1. 

Hence  3  and  —  1  are  the  ordinates  of  the  points  of  contact.    Then,  from  (1), 
if  fc  =  0  and  y  =  3,  we  have  x  =  6 ; 
if  A:  =  2  and  ?/  =  —  1,  we  have  x  =  —  Q. 

Hence,  if  A;  =  0,  the  point  of  contact  is  (6,  3) ; 

if  A;  =  2,  the  point  of  contact  is  (—  6,  —  1). 

The  points  of  contact  may  also  be  found  by  solving  each  of  equations  (3) 
with  the  given  equation. 

PROBLEMS 

1.  Determine  the  condition  for  tangency  of  the  loci  of  the  following 
equations. 

(a)  4x2+?/2-4x-8=0,  y  =  2x-\-k.  Ans.  k^ -\-2k -\1  =0. 

(b)  xy  +  X  -  6  =  0,  x  =  ky  -^-b.  Aiis.  fe2  +  14  A;  +  25  =  0. 

(c)  x2  —  2/2  =  o?^  y  —  kx.  Ans.  A;  =  ±  1. 

(d)  x2  +  2/2  =  ^2^  4  2/  -  3  X  =  4  A;.  Ans.  16  A:2  =  25  r'^. 

(e)  x2  +  ?/2  zz  r2,  2/ =  mx  +  6.  Ans.  (2?w6)2-4(l  +  m2)(62_^2)^0. 

(f).^  +  ^^  =  l,-+^.l.  ^n.  ^  +  ^  =  1. 

^  -^     a2     .  62        '  a      ^  a2  ^  ^2 

/  ,.  x2      2/2      1     a:  ,  2/      ,  .        a2      52 

(g)     —c.-7-.  —  ^i  —  +  -  =  1.  Ans. =1. 

^^'     a2      62        '  a      ^  a2      ^2 

X2         W2 

(h)*  —  +  ^-  =  1,  2/  =  mx  +  /3.  Ans.  ahn^  +  6^  -  /32  =  q. 

a2      6^ 

0)*  ^  -  r^  =  1'  y  =  mx  +  p.  Ans.  a'^m^  -  62  -  /32  =  0. 

a2      62 

(j)*  x2  4-  2/2  =  r2,  x  cos  w  +  2/  sin  w  —  p  =  0. 

^ns.  p2  _  y2  _  0. 

(k)*  2 x^  =  a2  -  +  ^  =  1.  Ans.  aB  =  2  a2. 

(1)*  x2  +  2/2  ^  ^2^  ^x  +  %  =  1.  Ans.  A^r'^  +  ^2^2  ^^  1. 

*  In  these  problems  it  is  assumed  that  the  constants  involved  are  not  zero. 


LINE  AND  CONIC  233 

2.  Find  the  equations  of  thie  tangents  to  the  following  conies  which  satisfy 
the  condition  indicated,  and  their  points  of  contact.  Verify  the  latter  approx- 
imately by  constructing  the  figure. 

(a)  ?/2  —  4 a;^  slope  =  i.  Ans.   x  —  2y  +  i  =  0. 

(b)  x2  +  2/2  ^  16,  slope  =  -  f  Ans.   4:X  +  Hy  ±20  =  0. 

(c)  9 ic2  +  16  ?/2  =  144,  slope  =  —  |.      Ans.    x  +  4  y  ±  4  VlO  =  0. 

(d)  x2  —  4 y^  =  36,  perpendicular  to  6x  —  4y  +  9  =  0. 

Ans.  2x  +  Sy  ±3Vl  =  0. 

(e)  x2+2?/2-x  +  ?/=0,  slope=-l.        Ans.  x-\-y  =  l,  2x  +  2y  +  l  =  0. 

(f)  x^=4:y,  passing  through  (0,-1).  Ans.  y  =±x  —  1. 

(g)  x^  =  8y,  passing  through  (0,2).    Ans.  None, 
(h)  4 x2  -  y2  =  16,  slope  =  2.                Ans.  y  =  2x. 

(i)  xy  +  y^  ~4:X-\-8y  =  0,  parallel  to  2x  -4y  =  7. 

Ans.   x  =  2y,  x-2y  +  48  =  0. 
(j)  4  x2  +  2/2  -  6  X  +  6  2/  =  0,  passing  through  (1,  1). 

Ans.   x-y=0,  19x  +  ll2/-30  =  0. 
(k)  x2  +  2  X2/  +  2/2  +  8  X  -  6  2/  =  0,  slope  =  f . 

Ans.   4  X  —  3  2/  =  0. 
(1),  x2  +  2x2/  -  4x  +  2  ?/  =  0,  slope  =  2. 

Ans.    y  =  2x,  2x  -y  -\-10  =  0. 

(m)  y^  =  2px,  slope  =  m.  Ans.    y  =  mx-\--^' 

2  m 

(n)  h^x'^  +  a22/2  =  a2&2,  slope  =  m.        Ans.    y  =  mx  ±  Vo^iM^. 
(o)  2xy  =  a2,  slope  =  m.  Ans.   y  =  mx  ±  a  V—  2  m. 

3.  Find  the  highest  and  lowest  points  of  the  conic 

(a)  x2  +  6  xy  +  9  y2  _  6  X  =:  0.  Ans.    Highest  (f ,  f). 

(b)  x2  -  2x2/  -  4x  -  4?/  -  8  =  0.         Ans.    (0,  -  2),  (-  4,  -  6). 

(c)  x2  -  2/2  -  4  X  +  8  2/  -  16  =  0.  Ans.    None. 

JTmt.  Find  the  points  of  contact  of  the  horizontal  tangents. 

93.  Tangent  in  terms  of  its  slope.  The  method  of  the  preced- 
ing section  for  finding  a  tangent  with  a  given  slope  may  be 
applied  to  general  equations  and  yield  formulas  for  the  equation 
of  a  tangent  in  terms  of  its  slope. 

Theorem  I.  The  equation  of  a  tangent  to  the  parabola  y'^  =  2px 
in  terms  of  its  slope  m  is 

(I)  y  =  -,nx+£^. 


234  ANALYTIC  GEOMETRY 

Proof.  Eliminating  x  from  y  =  mx  +  k  and  y^  =  2px,  we  obtain 

m?/^  —  2py  +  2pk  =  0. 
Hence  the  condition  for  tangency  is 

A  z=  (-  2py  -4m  (2pk)  =  0,       , 

whence  k  =  7^ — 

2m 

Substituting  in  y  =  mx  +  k,  we  obtain  (I).  q.e.d. 

In  like  manner  we  prove 

Theorem  II.    The  equation  of  a  tangent  in  terms  of  its  slope  m 
to  the 


circle  x^  -\-  y^  =  r^      is  y  =  moo  Ht  r  Vl  +  m^ ; 

ellipse        Px^  +  a'^y'^  =  a%'^  is  y  =  thx^  i  '\/a^ni^  +  b^ ; 
hyperbola  V^x^  —  a^y^  =  a^"^  is  y  =  mx  ±_  ^ahn^  —  b^. 


PROBLEMS 

1.  Find  the  equations  of  the  common  tangents  to  the  following  pairs  of 
conies.     Construct  the  figure  in  each  case. 

(a)  2/2  =  5x,  9x2  +  9?/2  ^  I6.  Ans.   9x  ±  12y  +  20  =  0. 

(b)  9  »2  +  16  y2  ^  144^  7  x2  -  32  2/2  =  224.  Ans.    ±x-y±5  =  0. 

(c)  x2  +  2/2  =  49,  x2  +  2/2  -  20  2/  +  99  =  0. 

Ans.    ±  4x  -  32/  +  35  =  0,  ±  3x  -  4?/  +  35  =  0. 

Hint.   Find  the  equations  of  a  tangent  to  each  conic  in  terms  of  its  slope  and  then 
determine  the  slope  so  that  the  two  lines  coincide  (Theorem  III,  p.  88). 

2.  Two  tangents,  one  tangent,  or  no  tangent  can  be  drawn  from  a  point 
Pi  (a^i,  Vl)  to  the  locus  of 

(a)  y^  =  2px  according  as  2/1^  —  2pxi  is  positive,  zero,  or  negative. 

(b)  &2x2  +  a^yi  =  a262  according  as  b'^Xi^  +  a^yi^  —  aW  is  positive,  zero, 
or  negative. 

(c)  &2x2  _  diyi  —  CJ252  according  as  Hh:,^  —  a'hfi^  —  a262  is  negative,  zero, 
or  positive. 

3.  Two  perpendicular  tangents  to 

(a)  a  parabola  intersect  on  the  directrix. 

(b)  an  ellipse  intersect  on  the  circle  x2  +2/2  =  a2  4.  jj\ 

(c)  an  hyperbola  intersect  on  the  circle  x^  ^- y^  =  a"^  —  &2. 


LINE  AND  CONIC 


235 


94.  The  equation  in  p.   In  the  following  sections  we  shall  suppose  that  the 
line  is  given  in  parametric  form  (Theorem  XV,  p.  124), 


(1) 


Cx  =  Xi  + 

\y  =  yi  + 


p  cos  a, 
P  cos  p. 


The  geometric  significance  of  these  equations  should  be  constantly  borne  in  mind. 
A  line  is  given  which  passes  through  P^  (x^,  y^)  and  whose  direction  cosines  (p.  123)  are 


cos  a  and  cos  /3  (or  whose  slope  is  m  ■■ 


sm  a     cos  /3 


by  (I),  p. 


The  point  {x,  y)  or 


cos  a     cos  a 

(a^i  +  p  cos  o,  2/i  +  p  cos  ^)  is  that  point  on  the  line  whose  directed  distance  from  P^  is  the 
variable  p. 


Suppose  the  conic  is  the  parabola   >■ 
(2)  2/2  -  7;9X  =  0. 

If  the  point  (xx  +  p  cos  a^y^-^p  cos  jS)  on  (1)  lies  on  (2),  then  (Corollary, 

p.  53) 

{yx  +  p  cos  i3)2  -  2 p  (xi  +  p  cos  a)  =  0, 
or 


(3) 


cos2  /3  •  /32  +  (2  2/1  cos  ^  -2p  cos  a)p  +  (yi^  -  2pxi)  =  0. 


This  equation  is  called  the  equation  in  p  for  the  parabola.  Its  roots, 
p\  and  /92,  are  the  directed  lengths  P1P2  and  P1P3  from  Pi  to  the  points 
of  intersection  of  the  line  and  parabola. 

For  p  is  the  distance  from  P^  to  the  point  {x^  +  p  cos  a,  t/j  4-  p  cos  )3);  and  when  p  satisfies 
equation  (3)  the  point  \X^  +  p  cos  a,yi  +  p  cos  /3)  lies  on  the  parabola. 

Hence 

Theorem  HI.    The  directed  distances  from  Pi  (xi,  2/1)  to  the  points  of  inter- 
iction  of  the  line 

X  =  Xi  +  /?  cos  a^  y  =  y\-\-  p  cos  ^ 

md  the  parabola  ij"^  =  2px  are  the  roots  of  the  equation  in  p, 

cos2  /3  •  p2  4.  (2  2/1  cos  /3  -  2p  cos  a)p  -\-  {tji^  -  2pxi)  =  0. 


236  ANALYTIC  GEOMETRY 

The  equation  in  p  for  any  conic  is  the  equation  whose  roots  are  the 
distances  from  a  point  Pi  to  the  points  of  intersection  of  tlie  conic  and  the 
line  through  Pi  whose  direction  angles  are  or  and  ^.  The  method  used  in 
proving  Theorem  III  is  general  and  justifies  the 

Rule  for  forming  the  equation  in  p  for  any  conic. 

Substitute  Xi  +  p  cos  a  for  x  and  yi  -i-  p  cos  /3  for  y  in  the  equation  of  the 
conic  and  arrange  the  result  according  to  powers  of  p. 

For  convenience  of  reference  we  state  the  following  theorems  which  are 
proved  by  this  Rule. 

Theorem  IV.    The  equation  in  pfor  the  central  conic  tP-x'^  ±  a'^y'^  —  a%'^  =  0  is 
(IV) 
(62  cos2  a  ±  a2  cos2  ^)p2  +  (2  hHx  cos  or  ±  2  a'^yx  cos  /3)p  +  (62xi2  ±  cc^y^^  -  a^b'^)  =  0. 

Theorem  V.    The  equation  in  p  for  the  locus  of 

Ax^  +  Bxy  +  (72/2  +  Dx  +  ^y  +  F=  0 
is 

(V)      {A  cos2  or  +  E  cos  a  cos  jS  +  C  cos2  ^)  ^2 

+  [(2  Axi  +  Byi  +  D)  cos  a  +  {Bxi  +  2  Cyi  +  E)  cos  /3]  p 

+  (^Xi2  +  Bxm  +  Cyt'  +  Dxi  +  Eyi  -h  F)  =  0* 

The  relative  position  of  the  line  (1)  and  a  conic  depends  upon  the  discrimi- 
nant of  the  equation  in  p.  For  according  as  the  roots  of  the  equation  in  p 
are  real  and  unequal,  real  and  equal,  or  imaginary  (Theorem  II,  p.  3),  the 
line  and  conic  will  intersect,  be  tangent,  or  not  meet  at  all. 

PROBLEMS 

1.  Find  the  equation  in  p  for  each  of  the  following  conies. 

{&)  xy  =  8.  (e)  2x^  +  xy  +  Sx-iy  =  0. 

(b)  x2  +  2/2  =  9.  {i)  x'^  +  2 xy  -^  y'^  -  4x  =  0. 

(c)  8x2  _  y2  ^16.  (g)  xy  +  4x  -  8y  -  3  =  0. 

(d)  x2  -  2/2  -f  4  X  -  6  y  =  0.  (h)  x2  +  4  xy  +  2/2  -  3  x  =  0. 

2.  What  can  be  said  of  the  coeflBcients  and  roots  of  the  equation  in  p 

(a)  if  Pi  (xi,  2/i)  lies  on  the  conic  ? 

(b)  if  the  line  is  tangent  to  the  conic  at  Pi  ? 

(c)  if  the  line  meets  the  conic  at  infinity  ? 

(d)  if  Pi  is  the  middle  point  of  the  chord  formed  by  the  line  ? 

*  Notice  that  the  coeflacient  of  p2  is  found  by  substituting  cos  a  for  x  and  cos  j8  for  y 
in  the  terms  of  the  second  degree  in  the  given  equation.  The  constant  term  is  found  by 
substituting  x\  for  x  and  y^  for  y  throughout  the  given  equation.  Compare  the  coeffi- 
cients of  cos  a  and  cos  /3  within  the  brackets  with  equations  (6)  and  (7),  p.  171. 


LINE  AND  CONIC 


237 


3.  Determine  the  relative  position  of  the  following  lines  and  conies  and 
construct  the  figures. 


(a)  ?/2  _  4  X  +  4  =  0 


(b)  4  xy  +  3  ?/2  -  4  X  +  4  ?/  -  16  =  0 


f  X  =  3  +  5  p, 
<  n   .    ,  Ans.    Secant. 


r 


(c)  4x2  +  92/2  -  40a; -72y  + 100  =  0 


(d)  3x2  +  x?/  -  4 ?/2  -  X  +  y  =  0 


Vio 


I  VIO 


Ans.   Tangent. 


1- 


V2 


Ans.    Do  not  meet. 

\y=-2-hlp. 

Ans.    Line  is  part  of  conic. 


(e)  4x2-9?/2  =  36 


=  2-|p, 

=  3  +   4p. 

Ans.  -  Secant  with  one  point  of  intersection  at  infinity. 


12/ 


95.  Tangents,  We  shall  show  how  to  find  the  equation  of  a  tangent  to  a 
conic  by  means  of  the  equation  in  p  by  considering  the  tangent  to  the  parab- 
ola 2/2  —  2_px  =  0  at  the  point  Pi  (xi,  2/i)-     Let 

(1)  X  =  Xi  + /)  cos  or,  2/ =  :yi  4- /9  cos /3 

be  any  secant  through  Pi  intersecting  the  parabola  at  Pq.  One  root  of  the 
equation  in  p  is  pi  =  P1P2  and  the  other  is  p2  =  0.  Hence  (III),  p.  235, 
becomes  (Case  I,  p.  4) 

cos2/3  •  p2  +  (2 2/1  COS  ^  -  2p  cos  a)p  =  0. 
[Or  tlie  constant  term  is  zero  by  the  Corollary,  p.  53.] 
When  P2  approaches  Pi  the  line  becomes  tangent  (p.  207),  and  as  pi 
becomes  zero  we  must  have  (Case  III,  p.  5) 

(2)  2  2/iCos  j8  —  2p  cos  a  =  0. 

This  is  the  condition  that  (1)  is  tangent  to  the 
parabola.  Solving  (1)  for  cos  a  and  cos  /3  and  substi- 
tuting in  (2),  we  obtain 

2yiy-2px-2  y^  +  2pxi  =  0. 
But  since  y-^  —  2  pxi  this  reduces  to 
2/12/  -  p  (x  +  xi)  =  0, 
which  is  the  form  given  in  Theorem  III,  p.  214. 


238 


ANALYTIC  GEOMETRY 


±  -  V—  1,  so  the  slopes  of 


96.  Asymptotic  directions  and  asymptotes.  If  the  coeflacient  of  p2  in 
the  equation  in  p  is  zero,  then  one  root  is  infinite  (Theorem  IV,  p.  15) ;  and 
hence  the  line  and  conic  have  one  point  of  intersection  at  an  infinite  distance 
from  Pi.     The  direction  of  such  a  line  is  called  an  asymptotic  direction. 

Theorem  VI.  The  asymptotic  directions  of  the  hyperbola  are  parallel  to  the 
asymptotes,  of  the  parabola  are  parallel  to  the  axis,  while  the  ellipse  has  no 
asymptotic  directions. 

Proof.    Set  the  coefficient  of  p^  in  the  equation  in  p  for  the  hyperbola 
[(IV),  p.  236]  equal  to  zero.     This  gives 
62  cos2  a  -a^  cos2  /3  =  0. 

cos/3  b 

.-.  m  = =±-' 

cos  a         a 

Therefore  the  slopes  of  the  asymptotic  direc- 
tions are  the  same  as  those  of  the  asymptotes 
[(5),  p.  190].  ^^ 

Similarly  for  the  parabola  m  = —  =  0,  so 

cos  a 

that  the  asymptotic  direction  is  parallel  to  the  axis. 

-,,       ,       „.         .     ,.,  cosiS  b 

For  the  ellipse,  m  like  manner,  m  = =  ±  - 

^    '  cos  a  a 

the  asymptotic  directions  are  imaginary  ;  that  is,'there 

are  no  asymptotic  directions.  q.e.d. 

Corollary.  Every  line  having  the  asymptotic  direction 
of  a  conic  intersects  the  conic  in  but  one  point  in  the 
finite  part  of  the  plane. 

If  both  roots  of  the  equation  in  p  become  infinite, 
the  line  is  said  to  be  "tangent  to  the  conic  at  infinity" 
and  is  called  an  asymptote.  Using  this  definition  of 
the  asymptotes,  we  have,  in  justification  of  the  prelimi- 
nary definition  on  p.  189,  the  following  theorem. 

Theorem  VII.     The  equation  of  the  asymptotes  of  the  hyperbola 
&2x2  -  a2?/2  =  am  is  62^2  -  a2y2  ^  0. 

Proof.  Both  roots  of  the  equation  in  p  for  the  hyperbola  [(IV),  p.  236] 
will  be  infinite  if  (Theorem  IV,  p.  15) 

62  cos2  a  —  a2  cos2  /3  =  0  and  2  62^1  cos  a:  —  2  a'^yi  cos  ^  =  0. 

From  the  first  equation,      cos  /3  =  ±  -  cos  a. 

Substituting  in  the  second,  we  get  bxi  T  ayi  =  0  as  the  condition  that  Pi 
should  lie  on  an  asymptote.  But  this  is  the  condition  that  Pi  should  lie  on 
one  of  the  lines  6x  =F  ay  =  0  or  62x2  —  a'^y'^  =  0.  Hence  this  equation  is  the 
equation  of  the  asymptotes.  q.e.d. 


LINE  AND  CONIC 


239 


The  method  of  the  proof  justifies  the 

Rule  for  finding  the  equation  of  the  asymptotes  of  any  hyperbola. 
First  step.    Derive  the  equation  in  p  (Rule,  p.  236). 
Second  step.    Set  the  coefficients  of  p'^  and  p  equal  to  zero. 
Third  step.    Eliminate  co8  a  and  cos  ^  from  these  equations  and  drop  the 
subscripts  on  Xi  and  yi. 


PROBLEMS 

1 .  Find  the  equations  of  the  tangents  to  the  following  conies  drawn  from 
the  points  indicated.  (The  method  of  section  95  can  be  applied  whether  Pi 
lies  on  the  conic  or  not.) 

(a)  xy  =  16,  (4,  4).  Ans.   x  +  y  =  S. 

(b)  x2  +  2xy  =  4,  (2,  0).  Ans.   x-\-y  =  2. 

(c)  x2  =  4  y,  (0,  -  1).                  Ans.  x2  -  (y  +  1)2  =  0,  or  x  =  ±{y  +  1). 

(d)  x2-32/2  +  2x+19  =  0,  (-1,  2).  Ans.   (x  + 3?/ -  6)(x  -  3y  +  7)  =  0. 

2.  Determine  the  slopes  of  the  asymptotic  directions  of  the  following 
conies. 

(a)  x2  -  xy  -  6  2/2  -  8  X  =  0.  Ans.  i,  -  |. 

(h)  xy-y^  +  4x-6  =  0.  Ans.  0,1. 

(c)  x2  +  4  x?/  +  4  2/2  _  2  X  =  0.  Ans.  -  ^. 

(d)  4  x2  +  xy  +  2/2  _  3  =:  0.  Ans.  None. 

(e)  9  x2  -  6  xy  +  y2  _  2  2/  +  5  =  0.  Ans.  3. 

(f)  x2  +  6xy  +  4y2  =  lo.  Ans.  -  i,  -  1. 

(g)  xy  +  Dx  +  Ey  +  F=0.  Ans.  0,  oo. 

3.  Determine  whether  the  loci  of  the  equations  in  problem  2  belong  to  the 
elliptic,  hyperbolic,  or  parabolic  type. 

4.  Find  the  equations  of  the  asymptotes  of  the  following  hyperbolas. 


(a)  xy  -  y2  +  2  X  =  0. 

(b)  2x2 -xy -4  =  0. 

(c)  x2-6xy  +  8y2  =  10. 

(d)  xy  -  4  X  -  3  y  =  0. 

(e)  2x2-7xy  +  3y2  =  14. 

(f)  x2  -  4  y2  +  2  X  +  8  y  =  0. 


Ans.  y  +  2  =  0,  x-y  +  2  =  0. 

Ans.  x  =  0,  2x  —  y  =  0. 

Ans.  x-4y=0,  x-2y  =  0. 

Ans.  X  =  3,  y  =  4. 

Ans.  2x-y=:0,  x-3y  =  0. 

Ans.  x-2y  +  3  =  0,  x  +  2y-l  =  0. 

5.  Find  the  equations  of  the  asymptotes  of  the  hyperbolas  (a),  (b),  (f), 
and  (g)  in  problem  2. 

Ans.   (a)  75x-25y  + 296  =  0,  50x  +  25y  -  184  =  0; 
(b)y  +  4  =  0,  x-y  +  4  =  0; 
(f)x  +  4y  =  0,  x  +  y  =  0; 
(g)  X  +  jE;  =  0,  y  +  D  =  0. 

6.  Prove  that  the  parabola  has  no  asymptotes. 


240  ANALYTIC  GEOMETRY 

7.  Show  that  the  asymptotic  directions  of  the  locus  of  Ax^  +  Bxy  +  Cy^ 
+  Dx -i-  Ey  +  F  =  0  sue  determined  by  the  locus  of  Ax^  +  Bxy  +  Cy^  =  0. 

8.  By  means  of  problem  7  show  that  the  locus  of  the  general  equation 
of  the  second  degree  belongs  to  the  hyperbolic,  parabolic,  or  elliptic  type 
according  a,s  A  =  B^  -  4:  AC  is  positive,  zero,  or  negative. 

9.  Show  how  to  determine  the  direction  of  the  axis  of  any  parabola  by 
means  of  problems  7  and  8. 

97.  Centers.    The  problem  of  this  section  is  to  determine  the  center  of 

symmetry,  if  there  is  a  center,  of  the  locus  of 

(1)  Ax'^  +  Bxy  +  Cy^ -\- Dx  +  Ey  +  F  =  0. 

That  is,  we  seek  a  point  Pi{Xi,'yi)  which  is  the  middle 

point  of  every  chord  of  (1)  drawn  through  it. 

If  Pi  is  the  middle  point  of  the  chord  P2P3  formed  by 

the  line 

x  =  Xi-\-  pcosa,    y  =  yi  +  p  cos  jS, 

"^  then  the  roots  of  the  equation  in  p  must  be  equal  numer- 

ically with  opposite  signs.     Hence  the  coefficient  of  p  in  (V),  p.  236,  must  be 
zero  (Case  II,  p.  4). 

(2)  .-.  (2  Axi  +  Byi  +  D)  cos  a  +  {Bxi  +  2  C?/i  +  E)  cos  ^  =  0. 

If  Pi  is  the  middle  point  of  every  chord  passing  through  it,  (2)  is  satisfied 
by  all  values  of  cos  a  and  cos  /3.     For  cos  /3  =  0  and  cos  a  =  0  we  get 

(3)  2  Axi  +  Byi  +  D  =  0,     Bxi  +  2  Cyi -{- E  =  0, 

and  if  equations  (3)  are  satisfied,  (2)  is  always  satisfied. 

We  can  solve  (3)  for  a  single  pair  of  values  of  Xi  and  2/1  (Theorem  IV, 

p.  90)  unless 

—  =  —,  OT  A  =  B^-AAC  =  Q, 
B       2C 

and  the  locus  of  (1)  will  have  a  single  center.     But  if  A  =  0  there  will  be 

no  center  unless  at  the  same  time  —  =  — ,  when  every  point  on  the  line 

2  C      E 
'2.  Ax  -^  By  -{-  B  =i  0  will  be  a  center. 

Hence  we  have 

Theorem  VIII.    The  locus  of 

Ax'^  +  Bxy  +  Cy^  +  Dx-}-  Ey  +  F=0 
will  have  a  single  center  of  symmetry  if  A  =  B^  —  4:  AC  is  not  zero.     If  A  =  0 

there  will  he  no  center  unless  —  =  — ,  when  all  of  the  points  on  a  line  will 
,         ,  2C      E 

oe  centers. 

Corollary.     The  center  will  be  the  point  of  intersection  of  the  lines 

2Ax-\-By  +  n  =  0,    Bjo^'^Cy  -\-E  =  0. 


LINE  AND  CONIC 


241 


If  the  locus  is  of  the  elliptic  or  hyperbolic  type  (p.  195),  there  will  be  a  single  center. 
But  if  the  locus  belongs  to  the  parabolic  type,  there  is  no  center  unless  the  locus  degen- 
erates. If  the  locus  is  a  pair  of  parallel  lines,  then  every  point  on  the  line  midway 
between  them  is  a  center. 

To  find  the  center  in  a  numerical  example  we  proceed  as  in  the  above 
proof  as  far  as  equations  (3)  and  then  solve  those  equations. 


Ans.  (0,  0). 
Ans.  None. 
Ans.    (-12,  -4). 


PROBLEMS 

1 .  Find  the  centers  of  the  following  conies. 

(a)  x2  +  xy  -  4  =  0. 

(b)  x'^  -2xy  +  y^  -4:X  =  0. 

(c)  xy-2y^  +  4x-4.y  =  0. 

(d)  jc2  _  8x2/  +  16 ?/2  +  2 X  -  8 2/  -  3  =  0. 

Ans.    Any  point  of  the  line  x  —  4y  +  l=0. 

(e)  x2  +  4xy  +  y2  -  8x  =  0.  Ans.    (-  |,  f). 

(f)  4x2  +  I2xy  +  92/2  _  2x  +  6  =  o.  Ans.    None. 

(g)  4x2  +  12x2/ +  92/2 -4x- 62/ -8  =  0. 

Ans.    Any  point  of  the  line  2x  +  32/  —  1  =  0. 

2.  If  all  the  coefficients  of  the  general  equation  of  the  second  degree 
except  B  are  constant,  and  if  B  varies  so  that  B^  —  iAC  approaches  zero, 
how  does  the  center  of  the  locus  behave  ? 

98.  Diameters.    The  locus  of  the  middle  points  of  a  system  of  parallel 
chords  of  a  curve  is  called  a  diameter  of  the  curve. 
Consider  the  ellipse 

62x2  +  a22/2  =  a262 
and  the  system  of  parallel  lines  whose 
direction  angles  are  a  and  /3.    The  para- 
metric equations  of  that  line  through 
Pi{xi,  2/1)  are  (Theorem  XV,  p.  124) 
jc  =  xi  +  p  cos  a,  2/  =  2/1  +  /o  cos  /S. 
If  Pi  is  the  middle  point  of  the  chord, 
then  the  roots  pi  =  P1P2  and  p2  =  P1P3 
of  the  equation  in  p  [(IV),  p.  236]  must 
be  equal  numerically  with  opposite  signs.     Hence  (Case  II, 
2  b'^Xi  cos  a  +  2  a^yi  cos  j8  =  0. 
cos/3 


p.  4) 


Dividing  by  2  cos  a  and  setting  m 


(p.  235),  we  get 


cos  a 
62xi  +  a'^myi  =  0 
as  the  condition  that  (Xi,  ^i)  is  the  middle  point  of  a  chord  whose  slope  is  m. 
This  is  the  condition  that  Pi  should  lie  on  the  line 
DD' :  62x  +  a^rny  =  0. 


242 


ANALYTIC  GEOMETRY 


Hence  we  have 

Theorem  IX.     The  diameter  of  the  ellipse 

62x2  +  a22/2  =  a262 
bisecting  all  chords  with  the  slope  m  is 
(IX)  b^a^  +  a^my  =  O. 

This  reasoning  may  be  applied  to  any  conic,  and  justifies  the 

Rule  for  deriving  the  equation  of  a  diameter  of  a  conic  bisecting  all  chords 
with  the  slope  m. 

First  step.    Derive  the  equation  in  p  {Rule,  p.  236). 

Second  step.    Set  the  coefficient  of  p  equal  to  zero. 

Third  step.    Replace  Xi  and  yi  by  x  and  y  respectively,  and by  m. 

The  result  is  the  required  equation. 

By  this  means  we  prove 

Theorem  X.    The  diameter  bisecting  all  chords  with  the  slope  m  of  the 
hyperbola  b'^x^  -  a^y-  =  a-b^  is  b^x  —  a'^my  =  O ; 


parabola 


2px  is 


my  =  p. 


Corollary.    All  the  diameters  of  the  parabola  are  parallel  to  its  axis,  and 
every  line  parallel  to  the  axis  is  a  diameter. 

Theorem  XI.    The  diameter  of  the  locus  of 

Ax'2  +  Bxy  +  Cy^  +  Dx  +  Ey  +  F=0 
bisecting  all  chords  of  slope  m  is 
(XI)  2Ax  +  By  +  n  +  m {Bx  ■\- ^'Cy  -\- E)  =  O. 

Corollary.    The  diameter  passes  through  the  center  if  the  locus  has  a  center, 
and  every  line  through  the  center  is  a  diameter. 

Hint.  Apply  the  Corollary,  p.  240,  and  Theorem  XIII,  p.  119. 


Cr 

LINE  AND  CONIC  243 

PROBLEMS 

1.  Find  the  equation  of  the  diameter  of  each  of  the  following  conies  which 
bisects  the  chords  with  the  given  slope  m. 

(a)  x2  -  4 2/2  =  16,  m  =  2.  Ans.  x-Sy  =  0. 

(b)2/2  — 4x,  wi  =  — -J.  Ans.  y  +  4  =  0. 

(c)  xy  =  6,  m  =  3.  Ans.  y  -{-Sx  =  0. 

(d)  x2  -  xy  —  8  =  0,  m  =  l.  Ans.  x  -  y  =  0. 

(e)  x2  -  4 y2  +  4 a;  -  16  =  0,  m=-l.  Ans.  x  +  4?/  +  2  =  0. 

(f)  xy +  2y^-4x-2y-\-e  =  0,m  =  i.  Ans.  2x  +  lly  -16  =  0. 

2.  Find  the  equation  of  that  diameter  of 

(a)  4 x2  +  9 2/2  =  36  passing  through  (3,  2).  Ans.  2x-3y  =  0. 

(b)  2/^  =  4x  passing  through  (2,  1).  Ans.  y  =  \. 

(c)  xy  =  8  passing  through  (—  2,  3).  ^ns.  3  x  +  2  y  =  0. 

(d)  x2  —  42/  +  6  =  0  passing  through  (3,  —  4).  Ans.  x  =  3. 

(e)  xy  —  2/2  4.  2 X  —  4  =  0  passing  through  (5,  2).  Ans.  4x  —  9?/  —  2  =  0. 

3 .  Find  the  slope  of  (XI)  if  ^  -  4  ^  C  =  0.    How  may  the  result  be  inter- 
preted by  means  of  problems  8  and  9,  p.  240  ? 

4.  What  relation  exists  between  m\  the  slope  of  (XI),  and  m  ? 

Ans.   2  Cmm'  +  5  (m  +  m')  +  2^  =  0. 

5.  What  does  the  result  of  problem  4  become  for 

62 

(a)  the  ellipse  62x2  4-  ahj^  =  a^b^  ?  -Ans.   mm'  = • 

a2 
62 

(b)  the  hyperbola  62x2  -  a2y2  =  a^^  ?  Ans.    mm'  =  — . 

a2 

(c)  the  parabola  y^  =  2px?  Ans.   m'  =  0. 

6.  By  means  of  problem  5  discuss  the  relative  directions  of  a  set  of 
parallel  chords  and  the  diameter  bisecting  them. 

7.  Find  the  equation  of  the  chord  of  the  locus  of 

(a)  x2  +  2/^  =  25  which  is  bisected  at  the  point  (2,  1). 

Ans.   2x  +  y  —  5  =  0. 

(b)  4x2  —  2/2  =  9  which  is  bisected  at  the  point  (4,  2). 

Ans.   8  X  -  y  -  30  =  0. 

(c)  xy  =  4  which  is  bisected  at  the  point  (5,  3). 

Ans.    3x  +  5y-  30  =  0. 

(d)  x2  —  xy  —  8  =  0  which  is  bisected  at  the  point  (4,  0). 

Ans.   2x-y-8  =  0. 


244 


ANALYTIC  GEOMETRY 


99.  Conjugate  diameters  of  central  conies.  In  every  system  of  parallel 
chords  of  a  central  conic  there  is  one  which  passes  through  the  center  and 
which  is  therefore  a  diameter  (Corollary  to  Theorem  XI).  This  diameter 
and  the  one  bisecting  the  chords  parallel  to  it  are  called  conjugate  diameters. 


The  ellipse 
Let  m  be  the  slope  of  a  diameter 
of  the  ellipse 

&2x2  +  a2y2  =  aW. 

From  Theorem  IX  the  slope  of 
the  conjugate  diameter  is  (Corollary 
I,  p.  86) 

62 

m'  = 

62 

.-.  mm  = 

a2 

Hence 

Theorem  xn.  If  m  and  m'  are  the 
slopes  of  two  conjugate  diameters  of 
the  ellipse  h^x^  +  a^^  =  a^b^,  then 

&2 


(XII) 


niTn'  =  — 


a^ 


Corollary.    Conjugate  diameters  of 
the  ellipse  lie  in  different  quadrants. 


For  tn  and  m'  have  opposite  signs  since 
tlieir  product  is  negative. 


The  hyperbola 

Let  m  be  the  slope  of  a  diameter 
of  the  hyperbola 

62x2  -  a22/2  =  a262. 

From  Theorem  X  the  slope  of  the 
conjugate  diameter  is  (Corollary  I, 
p.  86) 

m'  = 


.-.  mm' 


62 

62 
a2* 


Hence  ^ 

Theorem  Xm.  If  m  and  m'  are  the 
slopes  of  two  conjugate  diameters  of 
the  hyperbola  b'^x'^  —  a^y^  =  a^b%  then 


(XIII) 


rmnf  = 


62 


"%.           y' 

yf/ 

^V^V 

^,^ 

\v: 

PPf 

^V\ 

l^^^J^ 

A-                  aJ^^ 

\v     ^^ 

y^''^ 

"^^^nVV 

/^^ 

^^K 

v/            y' 

vS 

Corollary.  Conjugate  diameters  of 
the  hyperbola  lie  in  the  same  quad- 
rant, but  on  opposite  sides  of  the 
asymptotes. 

For  m  and  rw'-liave  the  same  sign  since 
their  product  is   positive,  and  if  one  is 

numerically  less  than  _ ,  the  other  must  be 

a         T 
numerically  greater  than  _  which  is  the 

a 
slope  of  one  asymptote  [(5),  p.  190]. 


LINE  AND  CONIC 


245 


The  ellipse 

The  length  of  a  diameter  of  the 
ellipse,  or  of  its  conjugate  diameter, 
is  that  part  of  the  line  included 
between  the  points  of  intersection 
of  the  line  and  the  ellipse. 

Construction.  To  construct  the 
diameter  conjugate  to  a  given  diam- 
eter AB,  draw  a  chord  EF  parallel 
to  AB,  and  then  draw  the  diameter 
CD  bisecting  EF, 


The  hyperbola 

The  length  of  that  one  of  two  con- 
jugate diameters  of  the  hyperbola 
which  does  not  meet  the  hyperbola 
is  defined  to  be  that  part  of  the  line 
included  between  the  branches  of 
the  conjugate  hyperbola  (p.  189). 

Construction.  To  construct  the 
diameter  conjugate  to  a  given  diam- 
eter AB,  draw  a  chord  EF  parallel 
to  AB,  and  then  draw  the  diameter 
CD  bisecting  EF. 


Theorem  XIV.  Given  a  point 
Pi  (^1?  Vi)  on  the  ellipse  b^x^  +  a^y^ 
=  a^b"^,  the  equation  of  the  diameter 
conjugate  to  the  diameter  through  Pi 
is 
(XIV)      b^acix  +  aHjxy  =  O. 

Proof.  The  diameter  through  Pi 
passes  through  the  origin  (Corollary, 
p.  242),  and  hence  its  slope  is  (Theo- 
rem V,  p.  35)  m  =  — .    Then,  from 

X\ 

(XII),  the  slope   of  the   conjugate 

y^x-i 

diameter  is  m'  = ^ —    The  equa- 

a'^yx 

tion  of  the  line  through  the  origin 

with  the  slope  m'  is  (Theorem  V, 

p.  95) 

62xi 

y  =  -^x, 
a^yi 

which  may  be  written  in  the  form 

(XIV).  Q.E.D. 

Corollary.  The  points  of  intersec- 
tion of  (XIV)  with  the  ellipse  are 

\       b       a  /         \  b  a  / 


Theorem  XV.  Given  a  point 
Pii^h  Vi)  on  the  hyperbola  6^x2  _  a^y2 
=  a^b^,  the  equation  of  the  diameter 
conjugate  to  the  diameter  through  Pi 
is 

(XV)     b^ociQc  —  ahjiy  =  O. 

Proof.  The  diameter  through  Pi 
passes  through  the  origin  (Corollary, 
p.  242),  and  hence  its  slope  is  (Theo- 
rem V,  p.  35)  m  =  —  -    Then,  from 

Xi 

(XIII),  the  slope  of  the  conjugate 

b^Xi 
diameter  is  m'  =  -s—  •      The  equa- 
a^yi 

tion  of  the  line  through  the  origin 

with   the  slope  m'  is  (Theorem  V, 

p.  95) 

b'^xi 

a^yi 

which  may  be  written  in  the  form 

(XV).  Q.E.D. 

Corollary.  The  points  of  intersec- 
tion of  (XV)  with  the  conjugate  hyper- 
bola are 

\  b       a  /  \     b  a  / 


These  are  found  by  the  Rule,  p.  76. 


These  are  found  by  the  Rule,  p.  76. 


246  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  "What  is  the  relation  between  the  slopes  of  conjugate  diameters  of 
the  equilateral  hyperbola  2  xy  =  a^  ?  Ans.    m  +  m'  =  0. 

2.  The  tangents  at  the  ends  of  a  diameter  of  (a)  an  ellipse,  (b)  an  hyper- 
bola, are  parallel  to  the  conjugate  diameter. 

3.  The  tangent  at  the  end  of  a  diameter  of  a  parabola  is  parallel  to  the 
chords  which  the  diameter  bisects. 

4.  The  sum  of  the  squares  of  two  conjugate  semi-diameters  of  an  ellipse 
equals  a^  +  62. 

Hint.  Let  P^  (ajj,  y-^  be  any  point  on  the  ellipse.  Find  the  squares  of  the  distances 
from  the  center  to  P^  and  to  one  of  the  points  in  the  Corollary  to  Theorem  XIV,  add,  and 
apply  the  Corollary,  p,  53. 

5.  The  difference  of  the  squares  of  two  conjugate  semi-diameters  of  an 
hyperbola  equals  a^  —  &2. 

Hint.   See  the  hint  to  problem  4. 

6.  The  angle   between  two   conjugate  diameters  is  sin-^^^,    where 
a'  and  h'  are  the  lengths  of  the  conjugate  semi-diameters.        ^ 

7.  Conjugate  diameters  of  an  equilateral  hyperbola  are  equal  in  length. 

8.  Conjugate  diameters  of  an  equilateral  hyperbola  are  equally  inclined 
to  the  asymptotes. 

9.  The  lines  joining  the  ends  of  conjugate  diameters  of  an  hyperbola  are 
parallel  to  one  asymptote  and  bisected  by  the  other. 

10.  The  product  of  the  focal  radii  (problems  8  and  9,  p.  194)  drawn  to 
any  point  on  (a)  an  ellipse,  (b)  an  hyperbola,  equals  the  square  of  the  semi- 
diameter  conjugate  to  the  diameter  drawn  through  that  point. 

11.  The  asymptotes  of  an  hyperbola  are  conjugate  diameters  of  an  ellipse 
which  has  the  same  axes  as  the  hyperbola. 

12.  Show  that  the  conjugate  diameters  of  the  ellipse  in  problem  11  are 
equal. 

MISCELLANEOUS   PROBLEMS 

1.  Find  the  condition  for  tangency  of 

(a)  2/2  =  2px  and  Ax  ^-  By  +  C  =  0. 

(b)  &2a;2  +  a2y2  -  a^p-  and  J.x  +  2^?/  +  O  =  0. 

(c)  62^2  _  a27/2  ^  ^262  and  Ax  ^  By  ^  C  =  0. 

2.  Find  the  points  on  each  of  the  conies  where  the  tangents  are  equally 
inclined  to  the  axes.     When  is  the  solution  impossible  ? 


LINE  AND  CONIC  247 

3.  Find  the  points  on  the  ellipse  where  the  tangents  are  parallel  to  the 
line  joining  the  positive  extremities  of  the  axes. 

4.  The  perpendicular  from  the  focus  of  a  parabola  to  a  tangent  intersects 
the  diameter  drawn  through  the  point  of  contact  on  the  directrix. 

6.  The  perpendicular  from  a  focus  of  a  central  conic  to  a  diameter  inter- 
sects the  conjugate  diameter  on  the  directrix. 

6.  Tangents  at  the  extremities  of  a  chord  of  a  parabola  intersect  on  the 
diameter  bisecting  that  chord. 

7.  Find  the  equation  of  (i)  the  ellipse,  (b)  the  hyperbola,  referred  to 
conjugate  diameters  as  axes  of  coordinates.     (See  problem  10,  p.  172.) 

8.  Given  the  equation  in  p  for  an  equation  of  the  second  degree;  what 
may  be  said  of  the  relative  positions  of  the  line,  the  conic,  and  the  point  Pi 

(a)  if  the  constant  term  is  zero  ? 

(b)  if  the  coefficient  of  p  is  zero  ? 

(c)  if  the  coefficient  of  p^  is  zero  ? 

(d)  if  the  coefficient  of  p  and  the  constant  term  are  zero  ? 

(e)  if  the  coefficients  of  p^  and  p  are  zero  ? 

(f)  if  the  coefficients  of  p^  and  p  and  the  constant  term  are  zero  ? 

(g)  if  the  discriminant  is  positive,  negative,  or  zero  ? 

9.  Tangents  to  an  hyperbola  at  the  extremities  of  conjugate  diameters 
intersect  on  the  asymptotes. 

10.  The  area  of  the  parallelogram  formed  by  tangents  at  the  extremities 
of  conjugate  diameters  of  (a)  an  ellipse,  (b)  an  hyperbola,  is  4  ab,  that  is,  it 
is  equal  to  the  area  of  the  rectangle  whose  sides  equal  the  axes. 

11.  The  diagonals  of  the  parallelogram  circumscribing  the  ellipse  in 
problem  10  are  conjugate  diameters. 

12.  Chords  drawn  from  a  point  on  (a)  an  ellipse,  (b)  an  hyperbola,  to 
the  extremities  of  a  diameter  are  parallel  to  a  pair  of  conjugate  diameters. 

13.  The  directrix  of  a  parabola  is  tangent  to  the  circle  described  on  any 
focal  chord  as  a  diameter. 

14.  The  tangent  at  the  vertex  of  a  parabola  is  tangent  to  the  circle 
described  on  any  focal  radius  as  a  diameter. 


CHAPTER  XI 
LOCI.     PARAMETRIC  EQUATIONS 

100.  The  first  fundamental  problem  (p.  53)  of  Analytic  Geom- 
etry is  to  find  the  equation  of  a  given  locus.  In  this  chapter  we 
shall  first  give  some  additional  problems  which  may  be  solved  by 
the  Rule  on  p.  53,  using  either  rectangular  or  polar  coordinates 
as  may  be  more  convenient.  We  shall  then  consider  two  classes 
of  loci  problems  which  are  not  readily  solved  by  that  Rule  and 
which  include  nearly  all  of  the  important  loci  occurring  in  Ele- 
mentary Analytic  Geometry.  '^ 

PROBLEMS 

It  is  expected  that  tlie  locus  in  each  problem  will  be  constructed  and  discussed  after 
its  equation  is  found. 

1.  The  base  of  a  triangle  is  fixed  in  length  and  position.  Find  the  locus 
of  the  opposite  vertex  if 

(a)  the  sum  of  the  other  sides  is  constant.  Arts.   An  ellipse. 

(b)  the  difference  of  the  other  sides  is  constant.     *    Ans.   An  hyperbola. 

(c)  one  base  angle  is  double  the  other.  Ans.    An  hyperbola. 

(d)  the  sum  of  the  base  angles  is  constant.  Ans.   A  circle. 

(e)  the  difference  of  the  base  angles  is  constant.         Ans.   A  conic. 

(f)  the  product  of  the  tangents  of  the  base  angles  is  constant. 

Ans.    A  conic. 

(g)  the  product  of  the  other  sides  is  equal  to  the  square  of  half  the  base. 

Ans.    A  lemniscate  (Ex.  2,  p.  152). 
(h)  the  median  to  one  of  the  other  sides  is  constant.        Ans.    A  circle. 

2.  Find  the  locus  of  a  point  the  sum  of  the  squares  of  whose  distances 
from  (a)  the  sides  of  a  square,  (b)  the  vertices  of  a  square,  is  constant. 

Ans.   A  circle  in  each  case. 

3.  Find  the  locus  of  a  point  such  that  the  ratio  of  the  square  of  its  dis- 
tance from  a  fixed  point  to  its  distance  from  a  fixed  line  is  constant. 

Ans.   A  circle. 
248 


LOCI 


249 


4.  Find  the  locus  of  a  point  such  that  the  ratio  of  its  distance  from  a 
fixed  point  Pi  (cci,  yi)  to  its  distance  from  a  given  line  Ax  +  By  +  C  =  0  is 
equal  to  a  constant  k. 

Ans.   {A^  +  ^2  _  ^2^2)  a;2  _  2  k'^ABxy  -\- {A^ -\- m  -  k^B^)  y^ 

-  2  {A^xi  +  B^i-{-k^AC)x-2 {A'^yi  +  B^y^  +  k^BC) y 

+  (Xl2  +  2/i2)  (^2  4.  J52)  _  ^2(72  =  0. 

5.  Find  the  locus  of  a  point  such  that  the  ratio  of  the  square  of  its  dis- 
tance from  a  fixed  line  to  its  distance  from  a  fixed  point  equals  a  constant  k. 

Ans.  x^  —  k-{x  —  p)2  —  k'^y'^  =  0  if  the  F-axis  is  the  fixed  line  and  the 
X-axis  passes  through  the  fixed  point,  p  being  the  distance  from  the  line 
to  the  point. 

6.  Find  the  locus  of  a  point  such  that 

(a)  its  radius  vector  is  proportional  to  its  vectorial  angle. 

Ans.    The  spiral  of  Archimedes,  p  =  a9. 

(b)  its  radius  vector  is  inversely  proportional  to  its  vectorial  angle. 

Ans.    The  hyperbolic  or  reciprocal  spiral,  pd  =  a. 

(c)  the  logarithm  of  its  radius  vector  is  proportional  to  its  vectorial  angle. 

Ans.    The  logarithmic  spiral,  logp  =  ad. 

(d)  the  square  of  its  radius  vector  is  inversely  proportional  to  its  vectorial 
angle.  Ans.    The  lituus,  p^d  =  a^. 

101.  Loci  defined  by  a  construction  and  a  given  curve.  Many 
important  loci  are  defined  as  the  locus  of  a  point  obtained  by  a 
given  construction  from  a  given  curve.  The  method  of  treatment 
of  such  loci  is  illustrated  by 

Ex.  1.  Find  the  locus  of  the  middle  points  of  the  chords  of  the  circle 
x2  +  y2  -=  25  which  pass  through  Pg  (3,  4). 

Solution.  Let  Pi(Xi,  yi)  be  any  point 
on  the  circle.  Then  a  point  P  (x,  y)  on 
the  locus  is  obtained  by  bisecting  P1P2. 
By  the  Corollary,  p.  39, 

X  =  l(Xi  +  3),   2/=l(yi  +  4). 

(1)     .-.  a:i=:2x-3,  y^  =  2y-4. 
Since  Pi  lies  on  the  circle  (Corollary, 

xi^  +  2/i2  =  25. 
Substituting  from  (1), 

(2  X  -  3)2  +  (2  ?y  -  4)2  =  26, 
ic2  +  2/2_3a;_4y_o. 


p.  53), 


r> 

K 

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ba: 

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<^  -?/: 

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0 

.> 

X 

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\ 

Wn 

g 

Vt) 

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-'1 

^ 

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- 

y 

r' 

250 


ANALYTIC  GEOMETRY 


As  this  equation  expresses  analytically  that  P(x,  y)  satisfies  the  given 
condition,  it  is  the  equation  of  the  locus. 

The  locus  is  easily  seen  to  be  a  circle  described  on  OP2  as  a  diameter, 
since  its  center  is  the  point  (|,  2)  and  its  radius  is  f  (Theorem  I,  p.  131). 

The  method  may  evidently  be  expressed  as  follows  : 

Rule  for  finding  the  equation  of  a  locus  defined  by  a  construction 
and  a  given  curve. 

First  step.  Find  expressions  for  the  coordinates  of  any  point 
P\  (cci,  2/1)  on  the  given  curve  in  terms  of  a  point  P  (x,  y)  on  the 
required  curve. 

Second  step.  Substitute  the  results  of  the  first  step  for  the  coordi- 
nates in  the  equation  of  the  given  curve  and  simplify.  The  result 
is  the  required  equation. 

This  Eule  may  also  be  applied  if  polar  coordinates  are  used 
instead  of  rectangular  coordinates. 

Ex.  2.  The  witch.  Find  the  equation  of  the  locus  of  a  point  P  constructed 
as  follows :  Let  OA  be  a  diameter  of  the  circle  x'^  -\-y^  —  2ay  =  0  and  let  any 
chord  OPi  of  the  circle  meet  the  tangent  at  ^  in  a  point  B.  Lines  drawn 
through  Pi  and  B  parallel  respectively  to  OX  and  OY  intersect  at  a  point  P 

on  the  required  locus. 

Solution.  First  step.  Let  {x,  y) 
be  the  coordinates  of  P  and  (xi,  yi) 
of  Pi.     Then  from  the  figure 

(1)  2/1  =  y- 

From  the  similar  triangles  OCPi 
and  PiPB  we  have 
xi  yi 


(2) 


00  _  CPi 
PiP  ~  PB 

OM-OC=x- 


X  —  Xi     2a  —  y 
For  OC  =  «i,  P^P  =  OM-  OC=x-x^,  CP^=  iji,  P£  =  MB- 

Solving  (1)  and  (2)  for  Xi  and  ?/i,  we  obtain 
(3) 


MP=2a-y. 


xy 
2a 


Second  step.    Substituting  from  (3)  in  the  equation  of  the  given  circle 
x2  +  ?/2  —  2  ay  =  0,  we  get 

x2'i/2 


+  y'' 


or 
(4) 


4a2 

x2y  =  4a2(2a 


2  ay  =  0, 

y)- 


LOCI 


251 


The  locus  of  this  equation  is  known  as  the  witch  of  Agnesi. 
Discussion  (p.  74).    1.  The  witch  does  not  pass  through  the  origin  (Theo- 
rem VI,  p.  73). 

2.  The  witch  is  symmetrical  with  respect  to  the  F-axis  (Theorem  V, 
p.  73). 

3.  Its  intercept  on  the  F-axis  is  2  a,  but  the  curve  does  not  meet  the 
X-axis  (Rule,  p.  73). 

4.  No  values  of  x  need  be  excluded,  but  all  values  of  y  must  be  excluded 
except  0<y<2a. 

For,  solving  (4)  for  x  and  y  (Rule,  p.  73), 

y  = ,  x  =  ±2a\\ ~- 

Hence  y  is  real  for  all  values  of  x.  But  the  fraction  under  the  radical  is  negative  if 
2/>2a  or  if  y  <0,  and  hence  these  values  must  be  excluded. 

5.  The  witch  extends  indefinitely  to  the  right  and  left  and  approaches 
nearer  and  nearer  to  the  X-axis. 

For  from  the  first  of  equations  (5),  as  x  increases  without  limit  y  decreases  and 
approaches  zero. 

Ex.  3,  The  conchoid.  Find  the  locus  of  a  point  P  constructed  as  follows  : 
Through  a  fixed  point  A  on  the  F-axis  a  line  is  drawn  cutting  the  X-axis  at 
Pi.     On  this  line  a  point  P  is  taken  so  that  PiP  =  ±  6,  where  6  is  a  constant. 


(5) 


Solution.    First  step.    Use  polar  coordinates,  taking  A  for  the  pole  and 
A  ¥  for  the  polar  axis.     Then  it  AO  =  a  the  equation  of  XX'  is 
(6)  p  =  a  sec  d. 

For  the  equation  of  a  line  perpendicular  to  the  polar  axis  has  the  form  (p.  156) 

C  C 

pA  cos  0  +  C  =  0,  or  p  = sec  6,  and  its  intercept  on  the  polar  axis  is 

A  A 

If  the  coordinates  of  P  are  (/?,  6)  and  of  Pi  are  {pi,  6i),  then  in  any  one  of 
the  figures  we  have  by  definition 

AP  =  ^Pi  ±  b. 
.'.  0i  =  d,pi  =  pTh. 


252  ANALYTIC  GEOMETRY 

Second  step.    Substituting  in  (6),  we  obtain 

(7)  p  =  a sec d  ±h. 

The  locus  of  this  equation  is  called  the  conchoid  of  Nicomedes.     It  has  three 
distinct  forms  according  as  a  rs  greater,  equal  to,  or  less  than  6. 

Discussion  (p.  151).    1.  The  intercepts  on  the  polar  axis  AY  are  a+  6 
and  a  —  b. 

The  pole  also  will  lie  on  the  conchoid  if  a  sec  6±b  =  0or9  =  sec- 1  f  ±  -)• 

2.  The  conchoid  is  symmetrical  with  respect  to  the  polar  axis  AY. 
For  sec  (-  6)  =  sec  6  by  4,  p.  19. 

3.  The  conchoid  recedes  to  infinity  in  the  two  opposite  directions  perpen- 
dicular to  the  polar  axis  A  Y. 

For  if  0  =  -  or  ^ — ,  sec  0  =  oo  and  hence  p  =  oo . 

2  2  j^p 

4.  If  we  transform  to  rectangular  coordinates,  using  (2),  p.  155,  we  get 

(x2. +  2/2)(x-a)2  =  62ic2. 

A  is  now  the  origin  and  J.  Y  the  positive  axis  of  X.     To  translate  the  axes 

7t 

to  0  and  rotate  them  through we  set  (Theorem  III,  p.  164)  x  =  y'  -\-  a, 

y  —  —  x'.     We  thus  obtain 

(8)  x2?/2  =  (2/  +  a)2(62-2/2), 

which  is  the  equation  of  the  conchoid  referred  to  OX  and  OY. 

From  (8)  it  is  easily  seen  that  the  conchoid  approaches  nearer  and  nearer 
to  the  X-axis  as  it  recedes  from  the  origin. 


PROBLEMS 

1.  Find  the  locus  of  a  point  whose  ordinate  is  half  the  ordinate  of  a  point 
on  the  circle  x'^  ■\- y"^  =  64.  Ans.   The  ellipse  a:2  +  4  ?/2  =  64. 

2.  Find  the  locus  of  a  point  which  cuts  off  a  part  of  an  ordinate  of  the 
circle  x'^  +  2/2  =  a2  whose  ratio  to  the  whole  ordinate  is  h-.a. 

Ans.    The  ellipse  62^2  +  aHj'^  =  a^b^. 

3.  Find  the  locus  of  a  point  which  divides  an  ordinate  of  (a)  x^-\-y-  =  r^^ 
(b)  y2  _  2px,  (c)  2xy  =  a^  into  segments  whose  ratio  is  X. 

Ans.   (a)  X2x2  +  (1  +  X)2?/2  =  x2r2  ;  (b)  (1  +  \)^y2  =  2\^px; 
(c)  2  (1  +  X)  xy  ==  Xa2. 


LOCI  253 

4.  Find  the  locus  of  the  middle  points  of  the  chords  of  (a)  an  ellipse,  (b)  a 
parabola,  (c)  an  hyperbola  which  pass  through  a  fixed  point  P^  (X2,  2/2)  on 
the  curve. 

Ans.   A  conic  of  the  same  type  for  which  the  values  of  a  and  6  or  of  p  are 
half  the  values  of  those  constants  for  the  given  conic. 

5.  Lines  are  drawn  from  the  point  (0,  4)  to  the  hyperbola  x^  —  4^/2  =  16, 
Find  the  locus  of  the  points  which  divide  these  lines  in  the  ratio  1 :  2. 

Ans.   9  x2  -  36  y2  +  192  y  _  272  =  0. 

6.  Lines  drawn  from  the  focus  of  the  conic  (II),  p,  178,  are  extended  their 
own  lengths.     Find  the  locus  of  their  extremities. 

Ans.   A  conic  with  the  same  focus  and  eccentricity  whose  directrix  is 
x  =  -2p. 

7.  Lines  are  drawn  from  a  fixed  point  P2  {x^,  2/2)  to  (a)  the  line  Ax  +  By 
+  C  =  0,  (b)  the  parabola  y'^  =  2px,  (c)  the  central  conic  Ax^  +  By^  +  F=0. 
Find  the  locus  of  the  points  dividing  these  lines  in  the  ratio  X. 

Ans.  (a)  a  straight  line  parallel  to  the  given  line ; 

(b)  a  parabola  whose  axis  is  parallel  to  that  of  the  given  parabola ; 

(c)  a  central  conic  whose  axes  are  parallel  to  those  of  the  given  conic. 

8.  Find  the  locus  of  the  middle  points  of  chords  of  an  ellipse  which  join 
the  extremities  of  a  pair  of  conjugate  diameters. 

Ans.   2  62x2  +  2  a^y^  =  a^l^. 

9.  A  chord  OPi  of  the  circle  x"^  +  y^  -{■  ax  =  0  which  passes  through  the 
origin  is  extended  a  distance  PiP  =  a.     Find  the  locus  of  P. 


Ans.   The  cardioia  |  (^'  +  ^'  +  <^)^  =  ''Y  +  ^'); 
i,or  p  =  a{l  —  cose 


cos  6). 

10.  A  chord  OPi  of  the  circle  x^  +  7/^  —  2ax  =  0  meets  the  line  x  =  2  a  at 

a  point  A.     Find  the  locus  of  a  point  P  on  the  line  OPi  such  that  OP=PiA. 

f  y^(2  a x)  ^  x^ 

Ans.    The  cissoid  of  Diodes  v  «  '    ■ 

t  or  p  =  2asmd  tan  d. 

11.  Find  the  locus  of  the  point  P  in  problem  9  if  PiP  =  b. 

Ans.    The  lima5on  of  Pascal,  p  =  b  —  a  cos  d.     The  limaQon  has  three  dis- 
tinct forms  according  as  6  =  a. 

102.  Parametric  equations  of  a  curve.  Equations  (XY),  p.  124, 

X  =  Xi-\-  p  cos  a,   y  —  y^-{-  p  COS  (i, 

are  called  the  parametric  equations  of  the  straight  line  because 
they  give  the  values  of  the  coordinates  of  any  point  {x,  y)  on  the 
line  in  terms  of  a  single  variable  parameter  p.    In  general,  if  two 


254  ANALYTIC  GEOMETRY 

equations  give  the  values  of  the  coordinates  of  any  point  (x,  y) 
on  a  curve  in  terms  of  a  single  variable  parameter,  those  equations 
are  called  parametric  equations  of  the  curve. 

Ex.  1.  Find  parametric  equations  of  the  circle  whose  center  is  the  origin 
and  whose  radius  is  r. 

Solution.    Let  P  (»,  y)  be  any  point  on  the  circle  and  denote  Z  XOP  by  6. 
Then  from  the  figure 

(1)  x  =  r  cos  e^  y  —  r  sin  d. 

These  are  the  required  equations.  They  possess  two 
properties  analogous  to  those  of  the  equation  of  the 
locus  (p.  53). 

1.  Correspgnding  to  any  point  P  on  the  locus  there 
is  a  value  of  d  such  that  the  values  of  x  and  y  given 
by  (1)  are  the  coordinates  of  P. 
2.  Corresponding  to  every  value  of  Q  for  which  the  values  of  x  and  y  given 
by  (1)  are  real  numbers  there  is  a  point  P  (x,  y)  on  the  locus. 

The  parameter  in  the  parametric  equations  of  a  curve  may  be 
chosen  in  a  great  many  ways,  and  hence  the  parametric  equations 
of  the  same  curve  will  often  appear  in  very  different  forms. 

Thus  in  Ex.  1,  if  we  had  chosen  for  the  parameter  half  the  abscissa  of  P, 
denoting  it  by  ^,  then  ^  =  - '  and  from  the  figure  y  -  ±.  V7*2  —  x^,  whence  the 
parametric  equation  would  have  been  x  =  2  <,  y  =  ±  Vr^  —  4 1^. 

Rule  to  'plot  a  curve  whose  parametric  equations  are  given. 

First  step.  Assume  values  of  the  parameter  and  compute  the 
corresponding  values  of  x  and  y  from  the  given  equations. 

Second  step.  Plot  the  points  whose  coordinates  are  found  in  the 
first  step. 

Third  step.  If  the  points  are  numerous  enough  to  suggest  the 
general  shape  of  the  locus,  draw  a  smooth  curve  through  the  points. 

Ex.  2.    Plot  the  curve  whose  parametric  equations  are 

(2)  X  =  at^,     y  =  aH\ 

Solution.    Take  a  =  ^.     Then  equations  (2)  become 

(3)  x  =  it^     y  =  itK 


LOCI 


255 


First  step.    Assume  values  of  t  and  compute  x  and  y  from  (3) ,    For  example, 
if  t  =  2,  X  =  i22  =  2,  ?/  =  i  23  =  2.     This  gives  the  table  : 

I  FA 


t 

ic 

y 

0 

0 

0 

1 

.5 

.25 

2 

2 

2 

3 

4.5 

6.75 

etc. 

etc. 

etc. 

-  1 

.5 

-    .25 

-2 

2 

-2 

-3 

4.5 

-G.75 

etc. 

etc. 

etc. 

Second  step.    Plot  the  points  found. 

Third  step.    Draw  a  smooth  curve  through  these  points  as  in  the  figure. 

Rule  to  find  the  equation  of  a  curve  in  rectangular  coordinates 
whose  parametric  equations  are  given. 

Eliminate  the  parameter  from  the  parametric  equations. 

We  shall  justify  the  Kule  for  the  examples  in  this  section. 

In  Ex.  1,  if  we  square  each  of  the  equations  (1)  and  add,  we  obtain  (3,  p.  19) 

x2  +  y2  =  r^ 

which  is  the  equation  of  the  given  locus  (Corollary,  p.  58). 

In  Ex.  2,  if  we  cube  the  first  of  equations  (2)  and  square  the  second,  we  get 
a;3  =  aH%    y^  =  aH^. 
(4)  .-.  ?/2=aa;8. 

This  is  the  equation  of  the  semicubical  parabola  (p.  209) .    To  prove  that  (4)  is 
the  equation  of  the  curve  obtained  in  Ex.  2  we  mi^st  prove  two  things  (p.  53) : 
1.  The  coordinates  of  any  point  P\  {x\,  yi)  on  the  curve  satisfy  (4). 

ilf  Pi  (a;i,  yi)  is  on  (2),  then  (1,  Ex.  1)  there  is  a  value  ti  such  that 
(5)  xi  =  atr^,      yi  =  a'^tiK 

(6)  .-.  xi3  =  aHi^,  yi^  =  aHi^. 

(7)  .'.yi2=axiK 

Hence  xi  and  yi  satisfy  (4) . 
2.  Ifxi  and  y\  satisfy  (4),  then  Pi  (xi,  yi)  is  on  the  curve. 
For  if  (7)  is  true,  then  from  the  first  of  equations  (5)  we  obtain  a  value  ^i.    Sub- 
stituting xi  =  ati^  in  (7),  we  get  yi  =  aHi^.    Hence  X\  and  yi  are  given  by  (5),  and 


256 


ANALYTIC  GEOMETRY 


The  parametric  equations  of  a  curve  are  important  because  it 
is  sometimes  easy  to  express  the  coordinates  of  a  point  on  the 
locus  in  terms  of  a  parameter  when  it  is  otherwise  difficult  to 
obtain  the  equation  of  the  locus. 

Ex.  3.  The  cycloid.  Find  the  parametric  equations  of  the  locus  of  a  point 
P  on  a  circle  which  rolls  along  the  axis  of  x. 

3T 


(8) 


Solution.  Take  for  origin  a  point  0  at  which  the  moving  point  P  touched 
the  axis  of  x.  Let  a  be  the  radius  of  the  circle  and  denote  the  variable  angle 
ABP  by  d.     Then  (p.  18) 

PC  =  a  sin  d,  CB  =  a  cos  d. 

By  definition,  OA  =  arc^P  =  ad. 

For  an  arc  of  a  circle  equals  Its  radius  times  tlie  subtended  angle,  from  the  definition 
of  a  radian  (p.  19). 

Hence  from  the  figure,  if  (x,  y)  are  the  coordinates  of  P, 

x  =  OD  =  OA-PC  =  ae-asmd,  2j  =  DP  =  AB-CB  =  a-acosd. 

'X  =  aid  -sin  6), 

y  —  a{l  —  cos^). 

These  are  the  parametric  equations  of  the  cycloid. 

Discussion.  1.  The  cycloid  passes  through  the  origin,  for  if  ^  =  0, 
x  =  yz=0. 

2.  The  cycloid  is  symmetrical  with  respect  to  the  Y-axis  (Theorem  IV, 
p.  72,  and  4,  p.  19). 

3.  Its  intercepts  on  the  X-axis  are  2n7ta,  where  n  is  any  positive  or 
negative  integer,  or  zero. 

For,  from  the  second  of  equations  (8),  if  y/  =  0,  cos  9  =  1.  .'.  0  =  2mr  ;  and  hence  from 
the  first  of  equations  (8)  x  =  a  •  2mr. 

4.  The  cycloid  lies  entirely  between  the  lines  y  =  0  and  y  =  2a,  for 

-  1<C0S^<1. 

5.  The  cycloid  extends  indefinitely  to  the  right  and  left  and  consists  of 
parts  equal  to  OMN.  For  if  we  replace  d  in  (8)  by  2  mt  +  d,  y  is  unchanged 
while  x  is  increased  by  2  riTT  (compare  Ex.  2,  p.  81). 


LOCI 


257 


Ex.  4.  The  hypocycloid  of  four  cusps.  Find  the  parametric  equations  of 
the  locus  of  a  point  P  on  a  circle  which  rolls  on  the  inside  of  a  circle  of  four 
times  the  radius. 


Solution.  Take  the  center  of  the  fixed  circle  for  the  origin  and  let  the 
X-axis  pass  through  a  point  A  where  the  tracing  point  P  touched  the  large 
circle.     Denote  ZAOB  hy  6.     Then  Z  BCP  =  4d. 

For  by  hypothesis  arc  PB  =  arc  AB ;  and  from  the  definition  of  a  radian  (p.  19) 


arc  PB=^ZBCP,  arc  AB  = 

4 


ae. 


ZBCP  =  a9,  or  ZfiCP  =  4d. 


But 


whence 


Z  OCE  +  Z  ECP  +  Z  PCB  =  Tt. 
It 


...  _  _  ^  +  z  EGP 
2 

Z  ECP  =  " 


id=7t. 


S9. 


Then  (p.  18)  DC  =  CP  cos  (  -  -  3  ^ )  =  ^  sin  3  6,  (by  6,  p.  20) 

DP  =  CPsinf--30')  =  -cos3^,  (by  6,  p.  20) 

\2  /      4 

OE  =  OC  Gosd  =  —  cos^, 
4 


EC  =  OC  sin  0  =  -—  sin  d. 


258  ANALYTIC  GEOMETRY 

Hence 

(9) 


OE  +  DP  =  — cose  +  -  cos 3 d, 
4  4 


y  =  EC  -  DC  =  —  sin  ^ sin  3  ^. 

4  4 

But  from  8,  10,  and  14,  p.  20,  and  3,  p.  19, 

cos  3  <9  =  4  cos3  ^  -  3  cos  ^,  sin  3  ^  =  3  sin  ^  -  4  sin^  d. 

Substituting  in  (9),  we  get 
(10)  x  =  a  cos^  e,  y  =  a  sin^  e, 

wliich  are  the  parametric  equations  of  the  hypocycloid  of  four  cusps. 

Discussion.     1.  The  hypocycloid  of  four  cusps  does  not  pass  through  the 
origin  because  there  is  no  value  of  6  for  which  sin  6  =  cos  ^  =  0. 

2.  It  is  symmetrical  with  respect  to  both  axes  and  the  origin. 
For  if  6=  6i  gives  a  point  (a^i,  yi)  on  the  curve, 

then  e—Tt  —  di    gives  a  point  (—  x\,  y\)       on  the  curve, 

6=  It  -{-  d\     gives  a  point  {—  X\,  —  y\)  on  the  curve, 

and  6=27t  —  6i  gives  a  point  (xi,  —  yi)       on  the  curve. 

3.  Its  intercepts  on  both  axes  are  ±  a. 

.     For  if  ^  =  0,  — ,  TT,  — -,  then  either  a;  =  0  and  y  =  ±a,  OTy  =  0  and  x  =  ±a. 

4.  Values  of  x  and  y  numerically  greater  than  a  give  no  points  on  the 
curve  since  sin  d  and  cos  6  cannot  be  numerically  greater  than  1. 

5.  The  hypocycloid  is  therefore  a  closed  curve. 

PROBLEMS 

1.  Plot  and  discuss  the  following  parametric  equations.     Verify  the  dis- 
cussion by  finding  the  equation  in  rectangular  coordinates. 

/  X  2«-l  -«  +  3  ,  ^  ,^  2 

^'^"  =  TTT'^^TT¥~'  (e)x  =  4^,^.-. 

(b)  «  =  4  cos  0,  ?/  =  2  sin  0.  (f )  x  =  3  +  2  cos  d,  y  =  2smd  —  \. 

(c)  X  =  4  sec  6,  y  =  i  tan  6.  (g)  x  =  i  -f  4,  y  =  i  t^. 

(d)  x  =  t-S,  y  =  \tK  (h)  x=:a cos^ 6,  y  =  h sin^ d. 

2.  Find  the  equation  in  rectangular  coordinates  of  (a)  the  hypocycloid  of 
four  cusps  (Ex.  4),   (b)  the  cycloid  (Ex.  3). 

Ans.  (a)x'  +  y^=:a^;  (b)  x  =  avers-i-  —  V2  ay— 2/2,  where  vers  ^=1  — cos  ^, 
or  0  =  vers-i  (1  —  cos  6). 

3.  Show  that   x  = ,    y  = -,    the  fractions  having  the  same 

ct  +  d  ct  -\-  d 

denominator,  are  the  parametric  equations  of  a  straight  line.      Interpret 
the  meaning  of  Mf  (a)  c  =  d  =  l,  (b)  c  =  0. 


LOCI  259 

4.  If  the  denominators  of  the  fractions  in  problem  3  are  different,  then 
the  equations  given  are  the  parametric  equations  of  an  equilateral  hyperbola 
or  two  perpendicular  lines. 

6.  Find  the  parametric  equations  of  the  ellipse,  using  as  parameter  the 
eccentric  angle  0,  that  is,  the  angle  from  the  major  axis  to  the  radius  of  a  point 
on  the  auxiliary  circle,  p.  206,  which  has  the  same  abscissa  as  a  point  on  the" 
ellipse.     Discuss  the  equations.  Ans.   x  =  a  cos  <p,  y  =  &  sin  0. 

Hint.  Apply  problem  10,  p.  206. 

6.  Find  the  locus  of  a  point  Q  on  the  radius  BP  (Fig.,  p.  256)  if  BQ  =  b. 
Ans.    <     ~         ,         ^  '     The  locus  is  called  a  prolate  or  curtate  cycloid 

according  as  b  is  greater  or  less  than  a. 

7.  Given  a  string  wrapped  around  a  circle,  find  the  locus  of  the  end  of  the 
string  as  it  is  unwound. 

Hint.  Take  the  center  of  the  circle  for  origin  and  let  the  X-axis  pass  through  the  point  A 
on  which  the  end  of  the  string  rests.    If  the  string  is  unwound  to  a  point  B,  let  Z  A  OB  =  0. 


Ans.    The  involute  of  a  circle  <  •    „        „ 

{y  =  rsmd  —  rd 


sin^, 
cos^. 


8.   A  circle  of  radius  r  rolls  on  the  inside  of  a  circle  whose  radius  is  r'. 
Find  the  locus  of  a  point  on  the  moving  circle. 


Ans.   The  hypocycloid 


X  =  (r'  —  r)  cos  6  +  r  cos 
y  =  {r'  —  r)  sin  0  —  r  sin 


r 
r'  —  r 


r 
where  6  is  chosen  as  in  Ex.  4. 


9.  A  circle  of  radius  r  rolls  on  the  outside  of  a  circle  whose  radius  is  /. 
Find  the  locus  of  a  point  on  the  moving  circle. 


Ans.   The  epicycloid 


X  =  (r'+  r)  cosd  —  r  cos d, 

y  =  (r'  -{■  r)  sill  ^  —  r  sin 0, 

where  6  is  chosen  as  in  Ex.  4. 


103.  Loci  defined  by  the  points  of  intersection  of  systems  of 
curves.  If  two  systems  of  curves  involve  the  same  parameter,  the 
curves  of  the  systems  belonging  to  the  same  value  of  the  param- 
eter are  called  corresponding  curves.  Many  loci  are  defined,  or 
may  be  easily  regarded,  as  the  locus  of  the  points  of  intersection 
of  such  curves.     The  method  of  treatment  is  illustrated  by 


260 


ANALYTIC  GEOMETRY 


Ex.  1.  A  fixed  line  ^5  is  drawn  parallel  to  one  side  of  a  rectangle,  and  a 
variable  line  CB  parallel  to  the  other  side.  Find  the  locus  of  the  intersec- 
tion oi  AC  and  BD. 

Solution.  Let  the  lengths  of  the 
sides  be  a  and  6,  and  take  two  sides 
for  the  axes.  Then  the  vertices  are 
(0,  0),  (a,  0),  (0,  6),  and  (a,  6). 

Let  0A=  ^  and  OD  =  k.  Then 
the  coordinates  ofA,B,  C,  and  D  are 
respectively  (0,  /3),  (a,  /3),  (k,  6),  and 
(k,  0).  Hence  the  equations  of  ^C 
and  BD  are  respectively  (Theorem 
VII,  p.  97) 

(1)  {b-  p)x-ky  -}-  ^k  =  0. 

(2)  ^x  +  {k-  a)y  -  ^k  =  0. 

These  equations  represent  two  systems  of  lines  and  involve  the  same 
parameter  k.     To  each  value  of  k  corresponds  a  pair  of  lines  intersecting  in  a 


point  P  (x,  y)  on  the  locus, 
of  P  (Rule,  p.  76) 

(3) 


Solving  (1)  and  (2),  we  obtain  as  the  coordinates 
al3k 


y 


hk  +  ajS 
h^k 


ab 


bk  -]-  aj3  —  ab 

As  these  equations  express  x  and  y  in  terms  of  a  parameter  k,  they  are  the 
parametric  equations  of  the  locus.    The  equation  of  the  locus  may  be  obtained 
by  eliminating  k  (Rule,  p.  255).     To  do  this  multiply  the  first  of  equations  (3) 
by  6,  the  second  by  a,  and  subtract.     This  gives 
(4)  bx  —  ay  =  0. 

The  locus  is  therefore  a  diagonal  of  the  rectangle,  for  this  line  passes 
through  (0,  0)  and  (a,  b)  (Corollary,  p.  53). 

This  equation  might  also  be  obtained  by  adding  (1)  and  (2).  But  if  the 
elimination  were  difiicult  or  impossible,  we  would  content  ourselves  with  the 
parametric  equations  (3). 

The  method  of  solving  Ex.  1  may  be  summed  up  in  the 
Rule  to  find  the  equation  of  the  locus  of  the  joints  of  intersection 
of  corresponding  curves  of  tivo  systems^ 

First  step.  Find  the  equation  of  the  two*  systems  of  curves  defin- 
ing the  locus  in  terms  of  the  same  parameter. 

*  In  some  cases  the  definition  involves  but  one  system  of  curves.  In  such  cases  a 
second  system  which  passes  through  the  points  on  the  locus  may  frequently  he  found. 


LOCI 


261 


Second  stejj.  Solve  the  equations  of  the  systems  for  x  and  y  in 
terms  of  the  parameter.  This  gives  the  parametric  equations  of 
the  locus. 

Third  step.  Find  the  equation  of  the  locus  from  the  parametric 
equations  {Rule,  p.  255). 

If  only  the  parametric  equations  are  required,  the  third  step  may  he  omitted. 

If  only  the  equation  in  rectangular  coordinates  is  required,  it  may  he  ohtained 
hy  eliminating  the  parameter  from  the  equations  found  in  the  first  step,  for  the 
result  will  he  the  same  as  that  ohtained  hy  eliminating  the  parameter  from  the 
equations  found  in  the  second  step. 

Ex.  2.  Find  the  locus  of  the  points  of  intersection  of  two  perpendicular 
tangents  to  the  ellipse  h'^x^  +  a'^y'^  —  a^b'^  =  0. 

Solution.  First  step.  The  equation  of  a  tangent  in  terms  of  its  slope  t  is 
(Theorem  II,  p.  234) 

(4) 


2/  =  te  +  vaH-^  +  62. 


The  slope  of  the  tangent  perpendicu- 
lar to  (4)  is  (Theorem  VI,  p.  36)  -  i ; 

t 
and  hence  its  equation  is  (Theorem  II, 
p.  234)  

Second  step.  As  the  parametric 
equations  are  not  requii-ed,  this  step 
may  be  omitted. 

Third  step.    To  eliminate  t  from  (4)  and  (5)  we  write  them  in  the  forms 


tx-y  =  -  ■VaH'^  +  6-2, 
x-]-ty  =  Va2  +  62^2. 

Squaring  these  equations,  we  obtain 

<2x2  -  2  te2/  +  2/2  =  a2<2  +  h% 

ic2  +  2  txy  +  «22/2  =  a2  +  52^2. 
Adding,         (1  +  t^)  x2  +  (1  +  t^)  y^  =  (1  +  f^)  a2  +  (1  +  t^)  bl 

Dividing  by  1  +  t^,  we  get  the  required  equation, 
x2  +  y^  =  a2  +  62. 

The  locus  is  therefore  a  circle  whose  center  is  the  center  of  the  ellipse  and 
whose  radius  is  Va2  +  62,     It  is  called  the  director  circle. 


262  ANALYTIC  GEOMETRY 


PROBLEMS 


1.  Find  the  locus  of  the  intersections  of  perpendicular  tangents  to  (a)  the 
parabola,  (b)  the  hyperbola  (VI),  p.  185. 

Ans.    (a)  The  directrix ;  (b)  x"^  +  y^  =  a"^  -  W. 

2.  Find  the  locus  of  the  point  of  intersection  of  a  tangent  to  (a)  an  ellipse, 
(b)  a  parabola,  (c)  an  hyperbola  with  the  line  drawn  through  a  focus  perpen- 
dicular to  the  tangent. 

Ans.   (a)  x2  -f  2/2  :=  a2 ;  (b)  x  =  0 ;  (c)  x2  +  ?/2  =  a2. 

3.  Given  two  fixed  points  A  and  J5,  one  on  each  of  the  axes,  and  two  vari- 
able points  A'  and  B%  one  on  each  axis,  such  that  OA'  +  OB'  =  OA  +  05, 
find  the  locus  of  the  intersection  of  AB'  and  A'B.       Ans.  x  +  y  =  a  +  6. 

Hint.    Let  OA=a,OB=b,  and  OA'  =a  +  }c;  then  OB'  =  b-}c. 

4.  Find  the  locus  of  the  point  of  intersection  of  a  tangent  to  an  equi- 
lateral hyperbola  and  the  line  drawn  through  the  center  perpendicular  to 
that  tangent. 

Ans.    The  lemniscate  (Ex.  2,  p.  152)  (x2  +  2/2)2  ^  ^2  (x2  -  y^). 

6.  Find  the  locus  of  the  point  of  intersection  of  a  tangent  to  the  circle 
«^  +  2/^  +  2  ox  +  a2  —  62  —  0  and  the  line  drawn  through  the  origin  perpen- 
dicular to  it. 

Ans.    The  lima^on  (problem  11,  p.  253)  (x2  +  2/2  +  ax)2  =  b'^{x^  -f  2/2). 

6.  Find  the  locus  of  the  point  of  intersection  of  the  diagonals  of  a  trape- 
zoid formed  by  drawing  a  line  parallel  to  one  side  of  a  given  triangle. 

Ans.    A  median  of  the  triangle. 

7.  Find  the  locus  of  the  intersection  of  the  diagonals  of  a  rectangle 
inscribed  in  a  triangle. 

Ans.    The  line  joining  the  middle  points  of  the  base  and  altitude. 

8.  Find  the  locus  of  the  point  of  intersection  of  lines  drawn  through  the 
foci  of  an  ellipse  parallel  to  conjugate  diameters.  Ans.   An  ellipse. 

9.  Find  the  locus  of  the  foot  of  the  perpendicular  drawn  from  the  origin 
to  a  tangent  to  the  parabola  2/2  +  4  ax  +  4  a2  =  0. 

Ans.    The  strophoid  y^  =  x'^ . 

MISCELLANEOUS   PROBLEMS 

1.  Find  the  locus  of  the  center  of  a  circle  which 

(a)  has  a  given  radius  and  passes  through  a  given  point. 

(b)  passes  through  two  given  points. 

(c)  passes  through  a  given  point  and  is  tangent  to  a  given  line. 

(d)  is  tangent  to  a  given  circle  and  a  given  straight  lipe. 

(e)  is  tangent  to  a  given  circle  and  passes  through  a  given  point. 


LOCI  263 

2.  One  side  of  a  triangle  is  fixed  and  a  second  side  has  a  constant  length. 
Find  the  locus  of  the  middle  point  of  the  third  side. 

3.  The  extremities  of  a  straight  line  of  variable  length  rest  on  two  perpen- 
dicular lines.  Find  the  locus  of  its  middle  point  if  the  area  of  the  triangle 
formed  is  constant. 

4.  Find  the  locus  of  the  middle  point  of  that  part  of  a  line  through  a  fixed 
point  Pi  (iCi,  2/i)  which  is  included  between  two  perpendicular  lines. 

6.  A  line  of  fixed  length  moves  with  its  extremities  on  the  axes.  Show 
that  (a)  the  locus  of  any  point  on  the  line  is  an  ellipse ;  (b)  the  locus  of  the 
foot  of  the  perpendicular  drawn  from  the  origin  to  the  line  is  the  four-leaved 
rose  p  =  a  sin  2  ^. 

6.  Let  the  X-axis  cut  the  circle  x"^  +  y^  =  a^  at  A.  An  arc  ^jB  is  laid 
off  on  the  circle  equal  to  the  abscissa  of  a  point  on  the  parabola  y^  =  2-px, 
and  the  radius  OB  is  produced  a  distance  BP  equal  to  the  ordinate  of  that 
point.     Show  that  the  locus  of  P  is  the  parabolic  spiral  {p  —  a)^  =  2  apd. 

7.  The  cissoid  (problem  10,  p.  253)  is  the  locus  of  the  point  of  intersection 
of  a  tangent  to  the  parabola  y^  +  Sax  =  0  and  the  perpendicular  to  it  drawn 
through  the  origin. 

8.  Given  a  fixed  point  A  on  the  negative  part  of  the  X-axis  and  a  line 
through  A  meeting  the  F-axis  at  B.  On  either  side  of  B  a  length  BP  =  OB 
is  laid  off  on  AB.  Show  that  the  locus  of  P  is  the  strophoid  (problem  9, 
p.  262). 

9.  One  side  of  a  right  angle  ABC  passes  through  a  fixed  point  D,  while  a 
point  C  on  the  other  side  moves  along  a  fixed  line  EF  whose  distance  from  D 
equals  the  side  BC.  Prove  that  (a)  the  middle  point  of  BC  describes  a  cissoid 
(problem  10,  p.  253) ;  (b)  the  vertex  B  describes  a  strophoid  (problem  9,  p.  262). 


CHAPTER  XII 
THE  GENERAL  EQUATION  OF  THE  SECOND  DEGREE 

104.  If  the  general  equation  of  the  second  degree  (p.  132) 

Ax^  +  Bxy  +  C?/2  -f  Dx  +  ^2/  +  i^  =  0 

has  a  locus,  it  must  be  either  a  conic  or  a  degenerate  conic  (Theorem  XIII, 
p.  196).  The  method  of  determining  the  exact  nature  of  the  locus  was  to 
simplify  its  equation  by  a  transformation  of  coordinates,  a  process  which  is 
frequently  laborious.  The  principal  object  of  this  chapter  is  to  derive  rules 
by  which  the  exact  nature  of  the  locus  may  be  easily  ascertained.  In  this 
connection  the  expressions  ,   , 

H  =  ^  +  C, 
and  0  =  iACF  +  BDE  -  AE'^  -  CD^  -  FW- 

will  be  of  fundamental  importance. 

105.  Condition  for  a  degenerate  conic. 

Lemma  I.  If  an  equation  of  the  second  degree  is  transformed  by  a  transfor- 
mation of  coordinates,  then  the  left-hand  member  of  the  transformed  equation 
can  be  factored  when  and  only  when  the  left-hand  member  of  the  original  equa- 
tion can  be  factored* 

Proof  For  the  equations  of  a  transformation  of  coordinates  [(III),  p.  164] 
are  of  the  first  degree  when  solved  for  either  the  new  or  old  coordinates,  and 
hence  when  we  substitute  in  an  equation  whose  left-hand  member  is  factored 
the  result  is  an  equation  whose  left-hand  member  is  factored.  q.e.d. 

Lemma  IL  The  locus  of  an  equation  of  the  second  degree  is  a  degenerate  conic 
when  and  only  when  the  left-hand  member  of  its  equation  may  be  factored. 

Proof    By  a  transformation  of  coordinates  an  equation  of  the  second 
degree  may  be  reduced  to  one  of  the  forms 
(1)  ^x2  +  Cy2  +  iT  =  0,  C?/2  +  Dx  =  0,  Cy'^-\-F=  0, 

where  A,  C,  and  D  are  different  from  zero  (Theorem  XIII,  p.  196). 

*  "We  shall  say  that  the  left-hand  member  of  the  equation  can  be  factored  if  it  can  be 
written  as  the  product  of  two  factors  of  the^rs^  degree  in  x  and  y  (p.  17).    Hence 

x^-y'^={x-}-y){x-y) 
can  be  factored,  while  x^-  y  =  {x  +  Vy) {x  —  y/y) 

cannot  be  factored  in  this  sense. 

264 


GENERAL  EQUATION  OF  SECOND  DEGREE         265 

The  locus  of  the  first  of  equations  (1)  is  degenerate  when  and  only  when 
F=0,  the  locus  of  the  second  is  never  degenerate,  and  the  locus  of  the  third 
is  always  degenerate.*  Hence  the  locus  of  an  equation  of  the  second  degree 
is  degenerate  when  and  only  when  its  equation  may  be  reduced  to  one  of 
the  forms 

(2)  Ax^  +  Cy^  =  0,  Cy^  +  F  =  0. 
These  equations  may  be  written  in  the  forms 

{Vax  -^V~Cy)  {VAx  -  vCTcy)  =  0, 
( Vc ?/  +  V^^)  ( Vc  ?/  -  V^^)  =  0. 

Hence  equations  (2)  are  forms  of  equations  (1)  which  can  be  factored, 
and  they  are  evidently  the  only  such  forms. 

Hence  the  locus  of  an  equation  in  its  simplest  form  is  degenerate  when 
and  only  when  it  can  be  factored,  and  then  by  Lemma  I  the  same  is  true  of 
the  locus  of  any  equation  of  the  second  degree.  q.e.d. 

We  now  seek  the  conditions  which  the  coefficients  of 

(3)  Ax^  +  Bxy  +  Cy^-^Bx-\-Ey  +  F=0 

must  satisfy  in  order  that  the  left-hand  member  can  be  factored. 
Arranging  (3)  according  to  powers  of  x,  we  have 

(4)  Ax^  +  {By  +  D)x  +  Cy^-]-Ey  +  F=  0. 

Solving  for  x  (which  implies  that  A  is  not  zero),  we  may  write  the  left- 
hand  member  of  (4)  in  the  form  of  (6),  p.  3,  namely 

(5)  a(x      -(-^y  +  -P)+^^-4^Qz/^  +  (2i?i)-4^^)?/  +  J)-^-4^J'\ 

/         -{By  +  D)-V{B^-4AC)y^-\-{2BD-4,AF)y  +  D^-4AF\ 
V  2  A  /■ 

These  factors  will  be  of  the  first  degree  in  y  as  well  as  x  when  and  only 

when  the  quadratic  in  y  under  the  radical  can  be  written  in  the  second  form 

of  (7),  p.  4,  which  can  be  done  when  and  only  when 

(6)  (2  J5D  -  4  AEy2  -4{B^-  4  AC)  (i)^  -  4  AF)  =  0. 
Clearing  parentheses  and  dividing  by  —  10  J.,  we  obtain 

(7)  AACF-h  BDE  -AE^  -  CD'^  -F&  =  0. 

The  left-hand  member  of  (7)  is  called  the  discriminant  of  (3)  and  is  denoted 
by  0.  Hence  the  left-hand  member  of  (3)  can  be  factored  (footnote,  p.  264) 
when  and  only  when  its  discriminant  is  zero.    Then  from  Lemma  II  we  have 

*  The  equation  Cy^  +  F=  0  has  no  locus  if  C  and  F  have  the  same  sign  (p.  196),  hut  we 
shall  speak  of  this  as  a  degenerate  case  to  distingviish  it  from  the  equation  Ax^+Cy^+F=0, 
which  has  no  locus  if  F^Q,  and  A,  C,  and  F  have  the  same  sign  (p.  195),  for  the  former 
equation  has  the  same  form  as  that  of  a  degenerate  parabola  (p.  196),  while  the  latter  has 
the  same  form  as  that  of  a  central  conic. 


266  ANALYTIC  GEOMETRY 

Theorem  I.    The  locus  of  an  equation  of  the  second  degree 
Ax^  -\-Bxy  -{-  Cy^  +  Dx  -\-  Ey  +  F  =  0 
is  degenerate  when  and  only  when  its  discriminant 

Q  =  4:ACF+  BDE  —  AE^  —  CD^  —  FB^ 

IS  zero.  ^i  —  ■^.^-    ■"    -     -       ^ 

PROBLEMS 

1.  If  J.  =  0,  then  0  =  BDE  -  CD^  -  FB'^.     Show  that,  the  locus  of 

■     Bxy  -\-  Cy'^  -\-  Dx -\-  Ey  -\-  F  =  0 
will  be  degenerate  when  and  only  when  0  =  0. 

Hint.   Arrange  the  given  equation  according  to  powers  of  y. 

2.  liA  =  C  =  0,  then  e  =  DE  -  FB,  after  dividing  by  B  which  we  sup- 
pose  is  not  zero.     Show  that  the  locus  of 

Bxy  +  Dx  +  Ey  -h  F=0  yp 

will  be  degenerate  when  and  only  when  0  =  0. 

Hint.   If  the  given  equation  can  be  factored,  the  factors  must  have  the  form 
{A'x-\-B'){C'y  +  D')  =  0, 
as  otherwise  their  product  would  contain  x^  or  y^. 

Multiply  these  factors,  find  the  conditions  that  their  product  should  have  the  same 
locus  as  the  given  equation,  and  eliminate  A',  B',  C\  and  D'. 

3.  Are  the  loci  of  the  following  equations  degenerate  or  non-degenerate  ? 

(a)  x2  —  2  xy  +  ?/2  -  2  2/  —  1  =  0.  Ans.  Non-degenerate. 

(b)  x2  +  2  xy  +  ?/2  +  X  +  ?/  -  2  =  0.  Ans.  Degenerate. 

(c)  x2  +  y2  -  4  X  +  2  ?/  +  5  =  0.  Ans.  Degenerate. 

(d)  x2  +  xy  +  2/2  +  2  X  +  3  y  -  3  =  0.  Ans.  Non-degenerate. 

(e)  xy  -I-  X  -  2/  +  7  =  0.  Ans.  Non-degenerate. 

(f )  x2  -I-  2  xy  -  2/  +  3  =  0.                             _  Ans.  Non-degenerate. 
{g)  xy  +  2x  -  y  -  2  =  0.  Ans.  Degenerate. 

4.  Find  the  real  values  of  k  for  which  the  loci  of  the  following  equations 
are  degenerate. 

(a)  A;x2  +  (1  _  fc)  2/2  -  (2  +  fc)  =  0.  Ans.    0,  1,  -  2. 

(b)  x2  +  (1  +  A:)  7/2  -  4  A:x  -  16  =  0.  Ans.    -  1. 

(c)  xy  +  k  (x2  -  2/2)  =  0.  Ans.    All  values. 

5.  Find  all  possible  cases  in  which  equation  (3),  p.  265,  has  no  locus. 

Hint.    Solve  for  x  and  apply  Theorem  III,  p.  11,  assuming  that  A  is  positive  and 
noticing  that  the  discriminant  of  the  quadratic  in  y  under  the  radical  is  - 16  ^0. 

Ans.    0  >  0,  A  <  0  ;  0  =  A  =  0,  1)2-4  AF<  0. 


GENERAL  EQUATION  OF  SECOND  DEGREE 


267 


106.  Degenerate  conies  of  a  system.    Let  the  equations  of  two  conies, 
degenerate  or  non-degenerate,  be 

Ci :  Aix^  -}-  Bixy  +  Ci?/2  +  DyX  -h  Eiy -{- Fi  =  0 
and  C2 :  ^2^2  +  Bzxy  +  C^y^  +  Ax  +  Ezy  +  Fg  =  0. 


(1) 
or 

(2) 


Then  the  equation 

Aix^  +  Bixy  +  Ci?/2  +  Dix  +  ^ly  +  Fi 

+  A;  (^2^2  +  B^xy  +  C^y^  +  DoX  +  ^22/  +  1^2)  =  0, 


(^1  +  kA2)x^  +  (5,  +  ^J52)X2/  +  (Ci  +  fcC2)y2 

+  (Z>i  +  JCD2)  X  +  (^1  +  A:^2)  y  +  (i^i  +  kF2)  =  0, 

where  A:  is  an  arbitrary  constant,  will  represent  a  system  of  conic  sections. 

If  Ci  and  C2  intersect,  all  the  conies  of  the  system  will  pass  through  their 
points  of  intersection. 

This  is  proved  as  in  the  case  of  straight  lines  (Theorem  XIII,  p.  119)  and  circles 
(Theorem  IV,  p.  140). 

Ex.  1.    Find    the  values  of  k  for  which  the  conies   belonging  to  the  sys- 
tem   a;2  4. 2/2  -  4  +  A:  (x^  -  2/2  -  1)  =  0    are 
degenerate. 

Solution.    The  givep  equation  may  he  writ- 
ten in  the  form 
(3)     (1  +  A;)  a:2  +  (1  -  k)  y^  -  (^  ^  k)  =  0. 

Its  discriminant  is 

e  =  -  4:{1  +  k)  {1  -  k)  (4:  +  k). 

If  the  locus  of  (3)  is  degenerate,  then  (Theo- 
rem I,  p.  266) 

e=^-4il  +  k)(l-k)i^-{-k)  =  0. 
.:  k  =  -l,  1,  or  -4. 

If  A;  =  —  1,  (3)  becomes  21/2  —  3=0,  or  y  =  ±  Vf. 
U  k=  1,  (3)  becomes  2  a;2  —  5  =  0,  or  x  =  ±  V|. 
If  A;  =  - 4,  (3)  becomes  3x^-5y^  =  0,  or  y  =  ±V^x. 


\ 

i .- 

\  1 

1 

/ 

\ 

\ 

/ 

^ 

\ 

^ 

T 

II 

/ 

■< 

s 

^^ 

'«j 

A 

^ 

Js 

fc— 

2^\ 

{— 

k^ 

-1 

/ 

iS 

y^ 

\ 

A 

' 

\ 

r 

^\ 

\ 

k= 

-1 

X 

■i' 

^^ 

J> 

vv 

/ 

y 

^ 

^ 

/ 

/ 

V 

< 

/ 

, 

\ 

\ 

/ 

"v 

In  each  case  the  locus  is  a  pair  of  lines. 

The  figure  shows  the  circle  a:2  -|- 1/2  _  4  =  0,  the  hyperbola  a;2  —  ?/2 
and  the  three  pairs  of  lines. 


1  =  0. 


Theorem  n.  In  mery  system  of  conies  whose  equation  has  the  form  (1)  there 
is  at  least  one  degenerate  conic  and.,  in  general,  there  cannot  be  more  than  three. 
These  are  obtained  by  substituting  for  k,  in  the  equation  of  the  system,  the  roots 
of  its  discriminant  0. 


268  ANALYTIC  GEOMETRY 

Proof.  The  discriminant  ©  of  (1),  when  set  equal  to  zero,  gives  an  equa- 
tion of  the  third  degree  in  k. 

For  each  term  in  ©  consists  of  the  product  of  three  of  the  coefficients  of  (2),  and  such 
products  will  contain  the  third  power  of  k. 

The  roots  of  this  cubic  equation  will  be  the  values  of  k  giving  the  degen- 
erate conies  of  the  system  (Theorem  I,  p.  266).  There  are,  therefore,  not 
more  than  three  values  of  k  for  which  the  locus  is  degenerate. 

In  a  special  case,  however,  all  of  the  coefficients  in  this  cubic  might  be  zero,  in  which 
case  the  locus  of  (2)  is  degenerate  for  all  values  of  k  (see  problem  4,  (c),  p.  266). 

Two  or  all  three  of  the  roots  might  be  equal,  and  hence  there  might  be 
but  two,  or  even  but  one,  degenerate  conic  in  the  system. 

Two  of  the  roots  might  be  imaginary  and  hence  could  not  be  used.  But 
one  of  them  must  be  real,*  and  hence  there  is  always  at  least  one  real  value 
of  k  for  which  the  locus  of  (1)  is  degenerate,  t  q.e.d. 

Systems  of  conies  defined  by  equations  of  the  form  (1)  are  classified  according  to  the 
nature  of  the  common  solutions  of  C^  and  Cj.  In  Algebra  it  is  shown  that  two  equations 
of  the  second  degree  have,  in  general,  four  pairs  of  common  solutions  for  x  and  y.  Hence 
five  cases  arise : 

1.  Four  distinct  pairs  of  solutions. 

2.  Two  pairs  are  identical  and  the  other  two  pairs  are  distinct. 

3.  Three  pairs  are  identical  and  the  fourth  pair  is  different. 

4.  Two  pairs  are  identical  and  the  other  two  pairs  are  also  identical. 

5.  All  four  pairs  are  identical. 

If  the  four  pairs  of  solutions  are  all  real,  then  these  five  cases  have  the  following 
geometrical  interpretation. 

1.  Ci  and  Cj  have  four  distinct  points  of  intersection.  All  the  conies  of  the  system 
pass  through  these  four  points.  There  are  three  degenerate  conies  in  the  system  [Ex.  1 
and  problem  1,  (a)]. 

2.  Cj  and  Cg  are  tangent  at  one  point  and  intersect  in  two  other  points.  All  the  conies 
of  the  system  are  tangent  at  the  first  point  and  pass  through  the  other  two  points.  There 
are  two  degenerate  conies  in  the  system  [problem  1,  (b)]. 

3.  Cj  and  Cj  are  tangent  at  one  point  and  intersect  in  a  second  point.  All  the  conies 
of  the  system  are  tangent  at  the  first  point  and  pass  through  the  second  point.  There  is 
but  one  degenerate  conic  in  the  system  [problem  1,  (c)]. 

4.  Ci  and  C^  are  bi-tangent,  that  is,  tangent  at  two  different  points.  All  of  the  conies 
of  the  system  are  tangent  at  these  two  points.  There  are  two  degenerate  conies  in  the 
system  [problem  1,  (d)]. 

5.  C-i  and  C^  are  tangent  at  one  point  and  do  not  intersect  elsewhere.  All  of  the 
conies  of  the  system  are  tangent  at  this  point.  There  is  but  one  degenerate  conic  in 
the  system  [problem  1,  (e)]. 

*  In  Algebra  it  is  shown  that  the  imaginary  roots  of  an  equation  with  real  coefficients 
must  enter  in  pairs.  Hence  if  the  degree  is  an  odd  number,  one  root,  at  least,  must  be 
real. 

t  It  is  tacitly  assumed,  as  is  true,  that  not  all  of  the  roots  of  the  discriminant,  when 
substituted  for  k,  give  equations  which  have  no  locus.  But  this  point  is  not  essential  for 
our  further  reasoning. 


GENERAL  EQUATION  OF  SECOND  DEGREE         269 

PROBLEMS 

1 .  Find  the  values  of  k  for  the  degenerate  conies  of  the  following  systems. 
Plot  Ci,  C2,  and  the  degenerate  conies. 

(a)  x2  +  2/2  _  16  +  A; (x2  +  92/2  _  36)  =  0.  Ans.   fc  =  -  1,  -\,  -  f . 

(b)  4x2  +  2^2  _  i6x  +  A; (x2  ^y^-^x)  =  0.       Ans.   A:  =  -  2,  -  2,  -  1. 

(c)  x2  4.  2x2/  +  22/2  +  8x  +  82/  +  A:(x2  +  22/2  +  82/)  =  0. 

Ans.   fc  =  —  1,  —  1,  —  1. 

(d)  x2  +  2/2  _  36  +  A;  (x2  +  4  2/2  -  36)  =  0.  Ans.    k  =  -l,  -  1,  -  1. 

(e)  x"^  +  y^  -  4:x  +  k{ix^  +  y^  -  4:x)  =  0.        Ans.   k  =  -l,  -  1,  -  1. 

2.  Find  the  points  of  intersection  of  Ci  and  C2  in  problem  1. 

Ans.   (a)  (fV6,iVl0),(fV6,-iVl0),(-|y6,^Vi0),(-|V6,-|Vu;). 

(b)  (0,  0),  (0,  0),  (I,  I  V2),  (f ,  -fV2). 

(c)  (0,  -  4),  (0,  -  4),  (0,  -  4),  (0,  0). 

(d)  (6,0),  (6,0),  (-6,0),  (-6,0). 

(e)  (0,  0),  (0,  0),  (0,  0),  (0,  0). 

3.  Discuss  the  following  systems  of  conies. 

(a)  x2  +  2/2  -  16  +  A;  (x2  -  4  y2  +  16)  =  0. 

(b)  x2  -  2  2/  +  4  +  A;  (x2  +  8  2/)  =  0. 

(c)  X2/ +  82/ +  8  +  A:(X2/ +  8)  =  0. 

(d)  x2  +  2  2/2  -  8  +  A;  (x2  +  2/2  -  4)  =  0. 

(e)  x2  +  6  2/  +  9  +  A;  (x2  +  6  2/)  =  0. 

(f)  2j^-4.x-\-k{y^  +  ix)  =  0. 

(g)  x2  -  2/2  +  25  +  A;(x2  -f  2/2)  =  0. 
(h)  x2  -  2/2  +  A;  (x2  +  2/2)  =  0. 

(i)  2/2  -  4 X  -  16  +  fc(x2  -  2/2  +  8 X  +  16)  =  0. 
(j)  x2  -  2/2  +  A:(x2  -  42/2  -  3)  =  0. 

107.  Invariants  under  a  rotation  of  the  axes. 

Lemma  III.  If  the  axes  are  rotated  about  the  origin,  then  for  any  point  whose 
old  and  new  coordinates  are  respectively  (x,  y)  and  (x',  y')  we  have 

x2  +  2/2  =  x'-^  +  2/'2. 
Proof.    To  rotate  the  axes  through  an  angle  6  we  set  (Theorem  II,  p.  162) 

J  X  =:  x' cos  ^  —  2/' sin  ^, 
'^  \y  =  x' sin  d  -\-  y' cos 6. 

Then  x2  +  y2  _  ^x'  cos  d  -  y'  sin  9)^  +  (x'  sin  6  +  y'  cos  5)2 

=  x'2  (cos2  d  -f  sin2  5)  +  2/'2  (siu2  5  +  cos2  6) 
=  x'2  +  y-2^  (by  3,  p.  19) 

Q.E.D. 

The  lemma  is  evident  geometrically  since  x^  +  ?/2  and  x'^  +  if^  are  the  squares  of  the 
distance  from  the  point  to  the  origin  [(JY),  p.  31]  in  the  new  and  old  coordinates 
respectively. 


270  ANALYTIC  GEOMETRY 

"We  are  considering  in  this  chapter  the  equation 

(2)  Ax^  +  Bxy  +  Cy^  -\-  Dx -\-  Ey  +  F  =  0. 

If  we  substitute  from  (1)  without  simplifying  the  result,  we  obtain  an 
equation  of  the  form 

(3)  A'x'-2  4-  B'xY  +  Cy^  +  D'x'  +  EY  +  F  =  0,  ^ 
which  has  the  same  constant  term  (Corollary,  p.  170). 

Consider  the  system  of  conies 

(4)  Ax^  +  Bxy  +  Cy^  +  Dx  +  Ey  +  F  +  k{x'^  -^  y^)  =  0. 

If  the  axes  be  rotated  by  substituting  from  (1),  the  equation  of  the  system 

becomes 

(6)  ^ V2  +  B'xY  +  CY^  +  D'x'  +  EY  -\-F+k (x'2  +  y'^)  =  0. 

For  the  left-hand  member  of  (2)  becomes  the  left-hand  member  of  (3),  and  x^  +  y' 
becomes  a;'^  ^  y^2  \)y  Lemma  III. 

Denote  the  discriminants  of  (2),  (3),  (4),  and  (5)  by  0,  ©',  ©i,  and  0/ 
respectively.  The  locus  of  (4)  is  degenerate  when  and  only  when  (Theorem  I, 
p.  266) 

ei  =  i{A+k){C  +  k)F-\-  BDE  -  {A  +  k)E-2  -  {C -^  k) D'^  -  FB^  =  0, 
or 

(6)  AFk^-i-{4:AF-\-ACF-E^  -D'^)k  +  Q  =  Q* 

Similarly,  the  locus  of  (5)  is  degenerate  when  and  only  when 

(7)  4  Fk^  +  (4  A'F  +  4  C'F  -  E'^  -  B'^)  k  ^  Q'  =  0. 
The  roots  of  (6)  and  (7)  must  be  the  same. 

For  (4)  and  (5)  are  the  equations  of  the  same  conic  referred  to  different  axes.  Hence 
the  locus  of  either  equation  is  degenerate  if  the  locus  of  the  other  is  degenerate. 

Since  the  coefficients  of  A;^  in  (6)  and  (7)  are  equal,  the  other  coefficients 
must  also  be  equal.     Hence 

(8)  0'  =  0 

and  4:A'F  +^C'F-  E'^  -  D'^  =  4AF  +  4CF  -  E^  -  Z)2. 

An  expression  involving  the  coefficients  A,  B,  C,  D,  E,  and  F  whose  value 
remains  unchanged  when  the  axes  are  changed  is  called  an  invariant  of  the 
general  equation  of  the  second  degree  under  a  transformation  of  coordinates. 
It  is  assumed  in  this  definition  that  the  equation  in  the  new  coordinates  is  not 
simplified  by  multiplying  or  dividing  by  a  constant.  An  expression  involv- 
ing the  coordinates  which  remains  unchanged  when  the  equation  in  the  new 
coordinates  is  simplified  is  called  an  absolute  invariant.     Hence,  from  (8), 

*  This  quadratic  may  be  regarded  as  a  cubic  equation  with  one  infinite  root,  by  a 
theorem  analogous  to  Theorem  IV,  p.  15.  The  locus  of  (4)  for  k=co  is  x^  + y^  =  0,  which 
is  one  of  the  degenerate  conies  of  the  system. 


GENERAL  EQUATION  OF  SECOND  DEGREE  271 

Theorem  m.    The  discriminant  0  of  an  equation  of  the  second  degree  is 
invariant  under  a  rotation  of  the  axes. 

Corollary.  The  expression  ^  =  4-42^  +  4  CF  —  E'^  —  Ifi  is  invariant  under 
a  rotation  of  the  axes. 

Lemma  IV.    An  invariant  of 

(9)  Ax^  +  Bxij  +  C?/2  +  F  =  0 

under  a  rotation  of  the  axes  which  involves  only  A,  J?,  and  C  is  also  an 
invariant  of  (2). 

Proof.    Substituting  in  (9)  from  (1),  we  obtain 
A'x'^  +  B'xY  +  Cy^  -{-F  =  0, 
where  A',  B\  and  C  have  the  same  values  as  in  (3). 

For  when  we  substitute  from  (1)  in  Ax^  +  Bxy  +  Cy^  we  obtain  only  terms  in  x'^,  x'y', 
and  y'"^,  and  substituting  in  Dx  +  Ey  in  (2)  we  obtain  only  terms  in  x'  and  y'. 

Hence  an  expression  involving  only  A^  B,  and  C  will  be  an  invariant  of 
(2)  if  it  is  an  invariant  of  (9).  q.e.d. 

Theorem  IV.    The  expressions 

A  =  B^-4tAC,    1I  =  A  +  C, 

are  invariants  of  an  equation  of  the  second  degree  under  a  rotation  of  the  axes. 
Proof.   Consider  the  system 

(10)  Ax^  -\-Bxy  +  Cy^  +  F+k  {x^  +  7/2)  =  o. 

Rotating  the  axes,  this  equation  becomes 

(11)  A'x'^  +  B'xY  +  CY^  +  F  +  A:  (x'2  +  y-i^  ^  q. 

Denote  the  discriminants  of  (10)  and  (11)  by  0i  and  ©/.  Then  the  locus 
of  (10)  is  degenerate  when  and  only  when 

■  ei  =  4(A  -\-  k)  (C  +  k) F -  Fm  =  0 
or 

(12)  -  4  A;2  +  4  (^  +  C)  A:  -  (^  -  4  ^  C)  =  0. 
Similarly,  the  locus  of  (11)  is  degenerate  when  and  only  when 

(13)  0/  =  4  A:2  +  4  {A'  +C')k-  {B'^  -  4  A'C)  =  0. 

Since  (10)  and  (11)  have  the  same  locus,  (12)  and  (13)  have  the  same  roots. 
And  since  the  coefficients  of  A;2  in  (12)  and  (13)  are  equal,  the  remaining 
coefficients  are  equal.     Hence 

B'^-iA'C'  =  B^-^AC 
and  A''  +  C  =  A  -]-  C.  q.e.d. 


272  ANALYTIC  GEOMETRY 

Ex.  1.    Transform  x^-\-xy  +  x  —  2y  +  4:  =  0'bj  rotating  the  axes  through  — 
Compute  A,  H,  and  ©  for  the  given  and  required  equations. 

Solution.    In  the  given  equation 

A=1,B  =  1,C=0,D  =  1,E  =  -2,F=4:. 
.'.  H  =  1  +  0  =  1,  A  =  12  -  4  •  1  •  0  =  1, 

0=4-l-O-4  +  l-l(-2)-l(-2)2-O-12-4-12  =  -lO, 


To  rotate  the  axes  set  (Theorem  II,  p. 

162) 

X 

=  X' COS  — 
4 

-.^=^-f 

y  =  X'  &in-  + 

y'cos-  = 

x'  -\-  y' 

V2 

This  gives,  after  removing  parentheses  but  not 

clearing  of 

fractions, 

(14) 

V2 

3 

'  +  4  = 

:0. 

Here 

A  =  l, 

B  =  -l,  C-0,  D  = 

1 

E^- 

3 

V2' 

F  = 

4. 

.-.  A  =  i,  H  =  i,  e  =  -io. 

Hence  the  values  of  A,  H,  and  0  are  unchanged. 
But  if  we  clear  fractions  in  (14)  we  obtain 

V2a;'2  _  V2xV  -  cc' -  3y' +  4  V2  =:  0. 

For  this  equation         A  =  2,  H  =  V2,  0  =  -  20  V2. 

Hence  A,  H,  and  0  are  not  absolute  invariants  under  a  rotation  of  the  axes. 

H2  H3 

Theorem  V.    The  expressions  —  and  —  are  absolute  invariants  of  an  equa- 
A  0 

tion  of  the  second  degree  under  a  rotation  of  the  axes.* 

Proof.  The  given  expressions  are  invariants  because  A,  H,  and  0  are 
invariants.  To  show  that  they  are  absolute  invariants  we  must  prove  that 
their  values  are  unchanged  when  we  multiply 

(15)  Ax^  +  Bxy  +  Cy2  i-Dx  +  Ey-h  F=0 
by  a  constant.     Multiplying  (15)  by  k,  we  get 

(16)  kAx'^  +  kBxy  +  kCy^  +  kDx  +  kEy  -\- kF  =  0. 

Denote  the  invariants  of  (16)  by  A^-,  H^-,  and  Qk-     Then 

(17)  Ax,.  =  k^B^  -  4 kAkC  =  k2{B^-^AC)  =  k^A. 

(18)  Jlk  =  kA-\-kC  =  k{A  +  C)  =  kli. 

(19)  Sk  =  k^  {iACF  +  BDE  -  AE^  -  CD^  -  FB^)  =  k^B. 

*  The  proof  also  holds  for  a  translation  of  the  axes  after  Theorems  VI  and  VII  are 
proved. 


GENERAL  EQUATION  OF  SECOND  DEGREE         273 

Dividing  the  square  of  (18)  by  (17), 

Ak  ~  A 
Dividing  the  cube  of  (18)  by  (19), 

H2  H3 

Hence  —  and  —  are  absolute  invariants.  q.e.d. 

A  0 

PROBLEMS 

H2  H3 

1.  Compute  —  and  —  for  the  equations  in  problem  2,  p.  168,  and  also 

A  0 

for  their  answers. 

2.  The  values  of  A\  B\  and  C  in  (3),  p.  270,  are  respectively  the 
coefficients  of  x'2,  x'y\  and  y"^  in  (4),  p.  170,  Compute  the  values  of 
£'2  _  4  vl'C  and  A'  +  C  in  terras  of  ^,  B,  and  C 

rj 

3.  Show  that -===:z^  is  an  invariant  of  the  line  Ax  -i-  By  +  C  =  0 

±  V^2  +  ^ 

under  a  rotation  of  the  axes. 

4.  Show  that  — ^  -  is  an  invariant  of  the  line  Ax-\-By  -\-  C  =  0 
and  the  point  Pi  (iCi,  ^i)  under  a  rotation  of  the  axes. 


6.  Show  that  V(xi  -  Xz)^  +  (yi  -  2/2)2  is  an  invariant  of  the  points 
Pi{^u  yi)  and  P2(X2,  2/2)  under  a  rotation  of  the  axes. 

6.  Show  that     ^   ^ ^— ?  is  an  invariant  of  the  lines  Axx-\-  Biy  +  Ci  =  0 

AiA2-\-  B1B2 

and  A2X  +  B^y  +  C2  =  0  under  a  rotation  of  the  axes. 

7.  Interpret  geometrically  the  meaning  of  the  invariants  in  problems  3 
to  6. 

108.  Invariants  under  a  translation  of  the  axes. 
Theorem  VI.    The  expressions 

A  =  B^-4AC,     Ti=A-i-C 
are  invariants  of  an  equation  of  the  second  degree  under  a  translation  of  the  axes. 

Proof.  If  an  equation  of  the  second  degree  be  transformed  by  translating  the 
axes,  the  coefficients^,  B,  and  C  are  unchanged  (Corollary  I,  p.  171).  Hence 
any  expression  involving  these  letters,  as  A  or  H,  is  an  invariant.  q.e.d. 


274  ANALYTIC  GEOMETRY 

Lemma  V.  If  the  axes  are  translated  to  the  point  {h,  fc),  then  for  any  point 
P  whose  old  and  new  coordinates  are  respectively  (x,  y)  and  (x',  y')  we  have 

kx  —  hy  =  kx'  —  hy\ 
Proof.    To  translate  the  axes  we  set  (Theorem  I,  p.  160) 

x  =  x'  -\-h,    y  =  y'  -\-k. 

Then   '  kx  -  hy  :^k{x'  +  h)  -  h  {7/  +  k) 

=  kx'  —  hy'.  Q.E.D. 

The  Lemma  is  evident  geometrically  since  either  kx  -  hy  or  kx^  -  Tiy'  is  the  area  of  the 
triangle  whose  vertices  are  P  and  the  old  and  new  origins  [(VIII),  p.  42]. 

Theorem  VII.     The  discriminant  0  of  the  equation 
(!)  Ax^  +  Bxy  +  Cy^+Dx-{-Ey  +F=0 

is  an  invariant  under  a  translation  of  the  axes 

(2)  x  =  x'  +  h,     y  =  y'  +  k. 
Proof.    Consider  the  system 

(3)  Ax^  +  Bxy  +  Cy^ -\- Dx  +  Ey  -{- F  +  k'  {kx  -  hy)  =  0. 
Substituting  in  (3)  from  (2),  we  obtain 

(4)  Ax'^  +  BxY  +  Cy'^  +  D'x'  +  Wy'  +  F'  +  ¥  {k<c'  -  hy')  =  0. 
For  (1)  becomes  an  equation  of  the  form  (Corollary  I,  p.  171) 

(5)  Ax'^  +  Bx'y'  +  Cy'^  +  B'x'  +  E'y'  -\-F'  =  0, 

and  kx  —  hy  becomes  kx'  -  hy'  (Lemma  V). 

Denote  the  discriminants  of  (1)  and  (5)  by  0  and  0' ;  of  (3)  and  (4)  by  0i 
and  0i'.     If  the  locus  of  (3)  is  degenerate  (Theorem  I,  p.  266), 
01  =  ^ACF+B{p  +  k'k)  {E  -  k'h)-A  {E  -  k'h)^  -  C  {D -}-  k'k)^  -  FB^  =  0, 
or 

(6)  {Bhk  -Ah^-  Ck^)  k'^  +  {BEk  -  BDh  +  2AEh-2  CDk)  k'-\-e  =  0. 

Similarly,  the  locus  of  (4)  is  degenerate  if 

(7)  {Bhk  -  Ah^  -  Ck^)  k'^  +  {BE'k  - BD'h  +  2  AE'h  -  2  CB'k)k'  +  0'  =  0. 

Since  (6)  and  (7)  must  have  the  same  roots,  and  since  the  coefficients  of  k'^ 
are  equal,  then  the  remaining  coefficients  are  equal.     Hence 

0'  =  0.  Q.E.D. 

Since  any  transformation  of  coordinates  may  be  effected  by  a  rotation  and 
a  translation  of  the  axes,  the  results  of  Theorems  III,  IV,  VI,  and  VII  may 
be  embodied  in  a  single  theorem. 


GEKEKAL  EQUATION  OF  SECOKJ)  DEGREE         275 

Theorem  Vin.    If  the  equation 

Ax^  +  Bxy  +Cy^-i-Dx  +  Ey  +  F  =  0 
be  transformed  by  a  transformation  of  coordinates  into 

A'x"^  +  B'xY  +  CY'^  +  D'x'  +  E'y'  +  F'  =  0, 
then  A'  =  B'2-4A'C' =  :^ -4:AC  =  A, 

H'  =  ^'  +  C  =  ^  +  C  =  H, 
and  0'  =  4  A'C'F'  +  B'lTE'  -  A'E'^  -  C'TY^  -  F'B'^ 

=  4:ACF  +  BDE  -  AE^  -  CD^  -  FB^  =  0. 

That  is,  A,  H,  and  0  are  invariants  of  an  equation  of  the  second  degree  under 
any  transformation  of  coordinates. 

PROBLEMS 

1.  Compute  —  and  —  for  the  equations  in  problem  1,  p.  168,  and  the 

A  0 

answers. 

2.  Prove  that  the  expressions  in  problems  4  to  6,  p.  273,  are  invariant 
under  a  translation  of  the  axes  and  interpret  them  geometrically. 

3.  Prove  by  direct  substitution  that  ^  (Corollary,  p.  271)  is  invariant  under 
a  translation  of  the  axes  provided  that  A  =  0  =  0. 

109.  Nature  of  the  locus  of  an  equation  of  the  second  degree.  By  a  trans- 
formation of  coordinates  the  equation 

(1)  ^x2  +  Bxy  -^Cy^  +  Dx  +  Ey  +  F=0 
may  be  reduced*  to  one  of  the  forms  (Theorem  XIII,  p.  196) 

(I)  A'x"^  +  C'y'^  +  F'  =  0,  where  ^'  5^  0  and  C  ^^  0 ; 

(II)  CY'^  +  D'x'  =  0,  where  C"  7^  0  and  D'  5^  0 ; 

(III)  CY'2-i-F'  =  0,  where   C':^0. 

The  theory  of  invariants  enables  us  to  determine  to  which  one  of  these 
three  forms  a  given  equation  may  be  reduced  and  to  find  the  exact  nature  of 
the  locus  without  actually  effecting  the  transformation  of  coordinates. 

To  do  this  compute  the  numerical  values  of  A,  H,  and  0  for  the  given 
equation  (1).     We  have,  further, 

(2)  for     (I),  A'  =  -  4  A'C  7^  0,  W  =  A' +  C^  0'  =  4  A'C'F'; 

(3)  for    (II),  A'  =  0,  H'  =  C"  7i  0,      0'  =  -  C'D''^  5^  0  ;  ' 

(4)  for  (III),  A'  =  0,  H'  =  C  5^  0,     0'  =  0. 

But  in  each  case,  by  Theorem  VIII, 

(5)  A'  =  A,     H'  =  H,     0'  =  0. 

*  It  is  assumed  that  the  equation  is  not  multiplied  or  divided  by  a  constant  in  this 
reduction. 


276  ANALYTIC  GEOMETRY 

Hence,  t/  A  7^  0,  (1)  may  he  reduced  to  the  form  (I) ; 

i/  A  =  0  and  0  7^  0,  (1)  may  be  reduced  to  the  form  (II) ; 
and  i/  A  =  0  and  0  =  0,  (1)  may  he  reduced  to  the  form  (III). 

We  shall  discuss  these  three  cases  separately. 

Cask  I.    A  5^  0.    Substituting  from  (2)  in  (5),  we  get 

(6)  -  4:A'C'  =  A, 

(7)  ^'  +  C'  =  H, 

(8)  4A'C'F'  =  Q. 

Elliptic  type,  A  <  0.  Hyperbolic  type,  A  >  0. 

From  (6),  if  A<0,  A'  and  C  have  From  (6),  if  A>0,  A'  and  C  have 

the  same  signs  and  the  locus  belongs  opposite  signs  and  the  locus  belongs  to 

to  the  elliptic  type  (p.  195).  the  hyperbolic  type  (p.  195). 

From  (8),  if  0  5^  0,  then  F'  ^0  and  From  (8),  if  0  9^  0,  then  F'  ^0  and 

the  locus  is  an  ellipse  if  H  and  0  differ  the  locus  is  an  hyperbola. 
in  sign,  or  there  is  no  locus  if  H  and  0 
agree  in  sign.    For  A'  and  C"  have  the 
sign  of  H,  from  (7),  and  F'  has  the  sign 
of  0,  from  (8). 

From  (8),  if  0  =  0,  then  ii^=  0  and  From  (8),  if  0  =  0,  then  i^=:  0  and 

the  locus  is  a  point.  the  locus  is  a.  pair  of  intersecting  lines. 

The  values  of  A^,  C,  and  F^,  if  desired,  may  be  found  by  solving  (6),  (7),  and  (8). 

Case  II.    A  =  0  and  0^0.    The  locus  is  a  parabola  (p.  180). 
Substituting  from  (3)  in  (5),  we  get  C^  =  H  and  —  CD'-  =  ©,  from  wbicb  the  values  of 
C  and  D'  may  be  found  if  desired.  , 

Case  III.  A  =  0  and  0  =  0.  Substituting  from  (4)  in  (5),  we  obtain  the 
single  equation  C  =  H,  which  does  not  enable  us  to  compare  the  signs  of 
C"  and  F'  in  (III).  But  ^  =  4.AF  +^CF  -FT^  -  D^  is  invariant  under  a 
rotation  of  the  axes,  and  when  A  =  0  =  0,  ^  is  also  an  invariant  under  a 
translation  of  the  axes. 

For,  substituting  the  values  of  D',  E',  and  F'  given  by  (5),  p.  170,  and  setting  A'  =  A, 
B'=B,  C'=  C  (Corollary  I,  p.  171)  in 

^'  =iA'F'  +  ^  C'F'  -  E'-^  -  Z>'2, 
weget  ^'={4:CD-2BE)h  +  {^AE-2BD)k  +  ^. 

But  if  A  =  0  =  0,  then  2  BD  -4AE  =  0,  from  (6),  p.  265.  Multiplying  this  by  B  and  set- 
ting B2  =  'i  AC  {from  A  =  0),  we  have  iACD  -  2  ABE  =  0,  or  4  CD  -  2BE=  0.    Hence  ^'=  ^ 

For  (III)  we  have  ^  =  4  C'F\  and  hence 

4  C'F'  =  ^ 

Hence  (p.  196)        if  ^  <  0,  the  locus  is  two  parallel  lines  ; 
if  ^  =  0,  the  locus  is  a  single  line ; 
if  f  >  0,  there  is  no  locus. 

The  results  of  this  section  are  embodied  in 


GENERAL  EQUATION  OF  SECOND  DEGREE         277 

Theorem  IX.    The  nature  of  the  locus  of  the  equation 

Ax^  +  Bxy  +  Cy^  -i-  Dx  +  Ey  +  F  .=  0 
depends  upon  the  values  of  the  invariants 

A  =  B^-4:AC,  R  =  A-\-C, 
e  =  ^ACF-{-  BDE  -  AE^  -  CD^  -  FB^, 
and  ^  =  4:AF-\-^CF-E'2-D^, 

as  indicated  in  the  following  table. 


09^0 

Conic. 

A<0 

Ellipse,  if  H  and  0  differ  in  sign. 
No  locus,  if  H  and  0  agree  in  sign. 

A  =  0 

Parabola. 

A>0 

-    Hyperbola. 

0  =  0 

Degenerate 
Conic. 

A<0 

Point. 

A  =  0 

Two  parallel  lines,  if  ^  <  0. 
One  line,  if  ^  =  0. 
No  locus,  if  ^  >  0. 

A>0 

Two  intersecting  lines. 

PROBLEMS 


1 .  Find  the  exact  nature  of  the  locus  of 

(a)  a^2  ^  2xy  -h  2y^  -  6x  -  2y  +9  =  0. 

(b)  x^  -  2 xy  -\-  2 y'2  -  4:y  +  8  =  0. 

(c)  x^-\-6xy  +  9y^-j-2x-6y  =  0. 

(d)  x2  -  2  xy  -  2/2  +  8  X  -  6  =  0. 

(e)  4  x2  4-  9  2/2  +  4  X  +  1  =  0. 

(f)  4x2  +  4x7/ +  ?/2  + 4x4-2  2/ -48  =  0. 

(g)  4x2 -20x?/4- 25?/2  + 12x-30?/  +  9  =  0 
(h)  9x2 -12x?/  +  4?/2- 18x4-122/ +  34  =  0. 

(i)  3x2  -  lOxy  +  7  2/2 +  15x- 7  2/ -42  =  0. 

2.  Find  a^  and  b^,  orp,  for  the  following  conies  : 

(a)  x2  -  2  X2/  +  2/2  -  8  X  =  0. 

(b)  3x2  -  10x2/  +  32/3-8  =  0. 

(c)  5  x2  +  2  x?/  +  5  2/2  -  12  X  -  12  2/  =  0. 

Hint.  Compute  the  absolute  invariants  —  and  —  for  tlie  given  equation  and  for  that 

A  0 

one  of  the  typical  forms  (III),  p.  179,  (V)  and  (VI),  p.  185,  to  which  it  may  be  reduced. 
Equate  and  solve  for  a^  and  b^  or  for  jo. 


Ans. 

Ellipse. 

Ans. 

No  locus. 

Ans. 

Parabola, 

Ahs. 

Hyperbola. 

Ans. 

Point. 

Ans. 

Two  parallel  lines. 

Ans. 

One  line. 

Ans. 

No  locus. 

Ans. 

Intersecting  lines. 

Ans.   p  =  V2. 

Ans.   a2  =  l,  62:^4. 

Ans.   a2  =  3,  62  =  2. 

278  ANALYTIC  GEOMETllY 

3.  Show  that  A'  and  C  in  (I),  p.  275,  are  the  roots  of  the  quadratic 
4x2  —  4Hx  —  A  =  0  and  show  that  they  are  always  real.  When  will  they 
also  be  equal  ? 

110.  Equal  conies.  The  object  of  this  section  is  to  determine  when  two 
conies  whose  equations  are  given  are  equal.  The  solution  of  this  problem 
affords  a  further  application  of  the  theory  of  invariants. 

Theorem  X.    The  axes  of  a  non-degenerate  central  conic  whose  equation  is 
Ax^  +  Bxy  +  Cy^  -{-  DX  +  Ey  +  F=0 

H2  H3 

are  determined  by  the  values  of  the  absolute  invariants  —  and  — 

A  0 

Proof.   The  equation  of  a  central  conic  may  be  reduced  to  the  form  [(11), 

P-  187] 

^  +  t  =  i 
a       p 

The  absolute  invariants  of  this  equation  are 


H'2      \a      /3/        (a  +  i8)2 

A'             -  4             -  4  a^  ' 

0'            -  4           -  4  a2^ 

a/3 

a/3 

Hence  (Theorem  VIII,  p.  275) 

(a  +  ^)2  _  H2 

{a  +  (8)3  _  H3 

-  4  a^        A  ' 

-  4  a2/32       0  ' 

(1) 

H2  H^ 

where  —  and  —  are  known.     These  equations  can  be  solved  for  a  and  /3, 

A  0 

and  the  values  of  the  axes  determined  from  them  by  1  and  2,  p.  187,  and  the 
definition  of  the  axes  (p.  185).  q.e.d. 

Equations  (1)  may  be  solved  as  follows  : 
Dividing  the  second  by  the  first, 

(2)  ^±1  =  ^. 
Dividing  the  first  of  equations  (1)  by  (2), 

(3)  -         ^™ 


Dividing  (3)  by  (2), 


A2 

4  ©2 


Then,  by  Theorem  I,  p.  3,  a  and  /3  are  the  roots  of  the  quadratic  equation 

4K©        4  ©2 
(4)  x^  +  ^^x-^-^  =  0,    or    A3a;2  +  4AH©a;-4©2  =  0. 

A2  A3 

The  roots  of  (4)  are  always  real,  for  the  discriminant  is 

(4  AH©)2  -  4  A3  (-  4  ©2)  =  16  A2©2  (H^  +  A) 

=  16A2©2(^2  +  2^C+C2+52-4^0 
=  16A202[(^-C)2+iB2], 

which  is  always  positive  when  the  coefiicients  A,  B,  C,  D,  E,  and  F  are  real  numbers. 


GENERAL  EQUATION  OF  SECOND  DEGREE         279 

Theorem  XI.    The  value  of  p  for  a  parabola  whose  equation  is 
Ax^  +  Bxy  +Cy^  +  Dx-{-Ey  +  F=0 

is  determined  by  the  value  of  the  absolute  invariant 

0 


Proof    For  the  parabola 
we  have 

7/2  =  2px 

H'3               IS                       1 
0'          -  4l)2              4p2 

Hence  (Theorem  VIII) 

1        H3 
4p2  -  0  ' 

whence 

^  =  W-|3- 

Q.E.D. 

As  the  value  of  p  is  always  a  real  number,  ©  and  H  must  have  opposite  signs.    This 
may  also  be  proved  from  the  values  of  ©  and  H  by  means  of  the  condition  A  =  0. 

Theorem  XII.    Two  non-degenerate  conies 

C:Ax'^  +  Bxy  +  Cy"^  +  Dx  +  Ey  +  F  =  0 
and  C  :  A'x^  +  B'xy  +  C'y^  +  D'x  +  E'y  +  F'  =  0 

are  equal  when  and  only  when 

A'   "*  A  '        0'   "  0  " 

Proof    If  the  conies  are  central  conies,  they  are  equal  when  and  only  when 
their  axes  are  equal.     But  the  axes  of  C  and  C  are  determined  in  the  same 

JJ2  JJ3  JJ'2  JJ'3 

manner  from  —  and  —  and  from  —  and  —  respectively  (Theorem  X). 
A  0  A'  0'        ^  "^  ^  ' 

Hence  the  axes  are  equal  when  and  only  when 

H'2      H2         ^     H'3      H3 

—  =  —    and     —  = 

A'        A  0'        0 

If  C  and  C"  are  parabolas,  they  are  equal  when  and  only  when  they  have 
le  same  value  of  p,  that  is  (Theorem  XI),  when  and  only  when 

= Q.E.D. 

0'  0 

111.  Conies  determined  by  five  conditions.   The  equation  of  any  conic 
las  the  form 

(1)  ^x2  +  Bxy  +  Cy^-\-  Dx  +  Ey  +  F=0, 

id  the  conic  is  completely  determined  if  five  of  the  coeflBcients  are  known 
terms  of  the  sixth.  Any  geometrical  condition  which  the  curve  must 
satisfy  gives  rise  to  an  equation  between  one  or  more  of  the  coeflBcients. 
Hence  five  conditions  will  determine  the  equation  of  a  conic.  The  locus  may 
be  degenerate,  or  there  may  be  no  locus,  which  would  mean  that  the  five 
conditions  are  inconsistent. 


280  ANALYTIC  GEOMETRY 

Rule  to  determine  the  equation  of  a  conic  which  satisfies  five  conditions. 
First  step.    Assume  that  the  equation  of  the  conic  is 

Ax^  +  Bxy  +  Cy"^ -{- Dx  +  Ey  -\- F  =  0. 

Second  step.    Find  five  equations  between  the  coefficients,  each  of  which 
expresses  that  the  conic  satisfies  one  of  the  given  conditions. 

Third  step.    Solve  these  equations  for  five  of  the  coefficients  in  terms  of  the 


Fourth  step.  Substitute  the  results  of  the  third  step  in  the  equation  in  the 
first  step  and  divide  out  the  remaining  coefficient.  The  result  is  the  required 
equation. 

PROBLEMS 

1.  Show  that  the  following  pairs  of  conies  are  equal  and  determine  the 
nature  of  the  conies. 

(a)  x^-iy'^-2x-16y  -U  =  0,  Sx^ +  10xy -\- Sy^  -  2  =  0. 

(b)  9x2  +  24xy  +  16y2_80x+602/=0,  x^-2xy  +  y^-4:V2x-4:V2y=0. 
(e)  x2  +  2/2  _  2x  -  8?/  -  8  =  0,  x'^-\-y^  +  6x-  lOy  +  9  =  0. 

(d)  2  x2  +  ?/2  -  12  X  +  10  2/  +  41  =  0,  17  x^  -12xy  -\-  22  y^  -  26  =  0. 

2.  Find  the  equations  of  the  conies  determined  by  the  following  condi- 
tions and  determine  the  nature  of  the  conic  in  each  ease. 

(a)  Passing  through  (0,  0),  (2,  0),  (0,  2),  (4,  2),  (2,  4). 

Ans.   x^  —  xy  +  y^  —  2x  —  2y  =  0. 

(b)  Passing  through  (0,  0),  (10,  0),  (5,  3)  and  symmetrical  to  the  A"-axis. 

Ans.    9x2  +  25^2  _  90x  =  0.  * 

(c)  Passing  through  (-  4,  0),  (0,  4),  (0,  -  4),  (5,  6)  if  A  =  0. 

Ans.    y^-4x-16  =  0. 

(d)  Passing  through  (0,  5),  (5,  0)  and  symmetrical  with  respect  to  both 
axes.  Ans.   x^  -\-  y^  —  25  =  0. 

(e)  Passing  through  (0,  0),  (2,  1),  (-  2,  4),  (-  4,  -  2),  (2,  -  4). 

Ans.    2  x2  -  3  X2/  -  2  ?/2  =  0. 

(f)  Passing  through  (0,  2),  (—  2,  0),  (2,  —  8)  and  symmetrical  with  respect 
to  the  origin.  Ans.   x^  -{■  ixy  +  y^  —  i  =  0. 

3.  Show  that,  in  general,  two  parabolas  may  be  constructed  which  pass 
through  four  given  points. 

4.  Find  the  parabolas  passing  through  the  following  points  and  construct 
the  figures. 

(a)  (0,  2),  (0,  -2),  (4,  0),  (-1,  0).     Ans.    x^  ±2xy -\- y^  -  Sx  -  4:  =  0. 

(b)  (2,0),  (0,-8),  (-2,0),  (0,2). 

Ans.    4  x2  ±  4  xy  +  ?/2  +  6  y  -  16  =  0. 

(c)  (0,1),  (0,-1),  (2,0),  (-1,0). 

Ans.    x^±ixy  +  iy^  —  x  —  2y  —  2  =  0. 


CHAPTER   XIII 

EUCLIDEAN  TRANSFORMATIONS  WITH  AN  APPLICATION 
TO   SIMILAR  CONICS 


112.  An  operation  which  replaces  a  given  figure  by  a  second  figure  in 
accordance  with  a  given  law  is  called  a  transformation.  If  a  transformation 
replaces  the  points  of  one  figure  by  the  points  of  a  second,  it  is  called  a  point 
transformation.  If  a  point  transformation  replaces  P(x,  y)  by  P'(x',  y'), 
then  the  equations  expressing  x'  and  y'  in  terms  of  x  and  y,  or  conversely, 
are  called  the  equations  of  the  transformation.  In  this  chapter  we  shall  con- 
sider the  transformations  which  replace  a  given  figure  by  one  equal  or  similar 
to  it.  They  are  called  Euclidean  transformations,  because  the  properties  of 
equal  and  similar  figures  are  studied  in  the  Elementary  Geometry  of  Euclid. 

113.  Equal  figures.    Two  figures  whose  corresponding  lines  and  angles 

are  equal  may  be  brought  into  coinci- 
dence and  are  therefore  equal.  Equal 
figures  in  the  same  plane  are  said  to  be 
congruent  if  the  corresponding  parts  are 
arranged  in  the  same  order,  and  sym- 
metrical if  they  are  arranged  in  the 
opposite  order.  Thus  the  triangles  ABC 
and  A'B'C  are  congruent,  and  either 
is  symmetrical  to  A"B"C'%  because  the 

directions  established  on  the  perimeters  by  the  corresponding  vertices  are  the 
same  (clockwise)  in  the  first  case  but  are  different  in  the  second  case. 

In  Plane  Geometry  we  do  not  study  symmetrical  figures  as  such. 
It  is  true  that  we  study  figures  which  are  symmetrical  with  respect 
to  a  point  or  with  respect  to  a  line.  But  it  should  be  noticed,  as 
is  seen  from  the  figures,  that  figures  which  are  symmetrical  with 
respect  to  &  point  are  congruent,  while  figures  which  are  symmetrical 

with  respect  to  a  line  are  sym- 
metrical in  the  sense  defined 
above. 

The  essential  distinction  be- 
tween congruent  and  symmet- 
rical figures  is  this  :  two  con- 
gruent figures  may  be  brought 
into  coincidence  by  moving  them 
around  in  the  plane,  but  before 
two  symmetrical  figures  can  be 
brought  into  coincidence  one  of  them  must  be  taken  out  of  the  plane  and  turned  over. 

281 


282 


ANALYTIC  GEOMETRY 


114.  Translations.  A  translation  is  the  transformation  which  moves  all 
points  of  a  figure  through  the  same  distance  in  the  same  direction.  Hence 
if  a  translation  replaces  any  point  P  by  P',  the  projections  of  PP'  on  the 
axes  will  be  constant. 

Theorem  I.  The  equations  of  a  translation  through  the  directed  length  whose 
projections  on  the  axes  are  respectively  h  and  k  are 


(I) 


00^  =  a^  -\-  h, 


Proof.    By  Theorem  III,  p.  31,  the  projections  of  PP^  on  the  axes  are 

respectively 

x'  -  X,     y'  -  y. 

Then,  by  hypothesis, 

x'—  x=  h,    y'—  y  =  k. 

Solving  for  x'  and  y%  we  obtain  (I).       q.e.d. 

If  we  solve  (I)  for  x  and  y  and  substitute  their 
values  in  the  equation  of  a  curve,  the  result  will 
X  evidently  be  the  equation  of  the  curve  after  it 
has  been  translated. 


If  P  is  the  origin  (0,  0),  then  P^  is  the  point  (h,  k).    If  we  solve  (I)  for  x  and  y,  we 
obtain 

x  =  x'-h,      y  =  y'-k. 

These  may  be  regarded  as  the  equations  for  translating  the  axes  to  a  new  origin 
{-h,  -k)  (Theorem  I,  p.  160). 


^>^ 


o' 


^.<^^^P) 


(1) 


ri 


(-hrk) 


(3) 


It  is  evident  that  the  relative  position  of  the  new  figure  and  the  old  axes  (Fig.  1)  is 
the  same  as  that  of  the  old  figure  and  the  new  axes  (Fig.  2). 

Hence  it  is.  immaterial  whether  we  regard  equations  (I)  as  the  equations  of  a  transla- 
tion of  a  figure  in  one  'direction  or  as  the  equations  of  a  translation  of  the  axes  in  the 
opposite  direction.  • 


116.  Rotations.  The  transformation  which  turns  all  points  through  the 
same  angle  about  a  given  point  0  is  called  a  rotation.  0  is  called  the  center 
of  the  rotation.     If  a  rotation  replaces  P  by  P',  then  OP'  =  OP. 


EUCLIDEAN  TRANSFORMATIONS 


283 


Theorem  n.    The  equations  of  a  rotation  about  the  origin  through  an  angle 
e  are 

=  acco89  —  y  sin  9^ 
05  sin  ^  +  y  cos  0. 


(H) 


Proof.   Let  the  polar  coordinates  of  P  be  {p,  <p).     Then,  by  definition, 
those  of  P'  are  {p,  <!>  +  6).    Hence  (Theorem 
I,  p.  155) 

x'  =  p  cos  (0  +  d) 

=  p  cos  <f>  COS  ^  —  ^  sin  0  sin  6 

(by  10,  p.  20) 
=  X  cos  ^  —  y  sin  ^, 
since  [(I),  p.  155] 

X  =  p  cos  ^,  y  =  p  sin  0. 
Similarly, 

y^  =  xshid  +  y  cosd.  q.e.d. 


Q 


/ 

/ 
/ 

Ad 


0 


If  we  solve  (II)  for  x  and  y,  we  get 

x  =  x'  cos  e  +  y^sine  =  x'  cos  (-  0)  -  y'  sin  (-  5), 
y  =  -x'9>m.Q  ^y'  cos  Q  —  x'  sin  (-  0)  +  y'  cos  (-  0). 


(by  4,  p.  19) 
(by  4,  p.  19) 


These  may  be  regarded  (Theorem  II,  p.  162)  as  the  equations  for  rotating  the  axes 
tlirough  an  angle  —  Q.  Hence  it  is  immaterial  whether  we  regard  equations  (II)  as  the 
equations  of  a  rotation  of  a,  figure  in  one  direction  or  of  the  axes  in  the  opposite  direction. 
This  should  be  illustrated  by  figures  analogous  to  Figs.  1  and  2,  p.  282. 


PROBLEMS 

1 .  Plot  the  following  curves,  translate  them  through  the  directed  length 
whose  projections  are  given,  and  find  the  equations  of  the  curves  in  their  new 
positions. 

(a)  y^  =  Ax,  h  =  -3,k  =  2.  Ans.   ?/2-4x-4?/-8  =  0. 

(b)  xy  =  6,h  =  2,k  =  -  2.  Ans.   xy  +  2x  -  2y  -  2  =  0. 

(c)  x2  +  92/2  =  25,  /i  =  0,  A:  =  |.  Ans.    x^  +  9y^  -  30y  =  0. 

2.  Plot  the  following  curves,  rotate  them  about  the  origin  through  the 
given  angle,  and  find  the  equations  of  the  curves  in  their  new  positions. 

Ans.    y^  -x^  =  16. 

Ans.   x^-^y'^+Sx  +  12  =  0.. 

Ans.   4  x2  +  2/2  4. 18  2/  =  0. 


(a)  xy  =  8,d 

It 
~  4 

(b)  x2  +  2/2  - 

8x  +  12  = 

0,  ^  = 

-  It. 

(c)  «2  +  42,2 

-18x  =  0, 

d  =  - 

It 
0 ' 

284  ANALYTIC  GEOMETRY 

3.  Translate  the  locus  of  x^  +  4  2/  =  0  through  a  distance  whose  projec- 
tions are  A  =  0,  k  =  —  4  and  then  rotate  it  about  the  origin  through  an  angle 

of-.  Ans.    2/2-4x  +  li3  =  0. 

2 

4.  Rotate  the  curve  in  problem  3  through  the  given  angle  and  then  trans- 
late it.  Ans.    2/2  _  4  X  +  8  2/  +  16  =  0. 

5.  Prove  from  equations  (II)  that  the  origin  is  unchanged  by  a  rotation, 
that  is,  that  the  origin  is  a  fixed  point. 

6.  Find  the  equations  of  the  straight  lines  which  are  unchanged  by  the 
translation  (I). 

Hint.  Translate  Ax  +  By  +  C=0  and  then  determine  A,  B,  and  C  so  that  this  line 
coincides  with  the  line  into  which  it  is  translated  by  Theorem  III,  p.  88. 

Ans.    kx  —  hy  —  0. 

7.  Find  the  equations  of  all  circles  which  are  unchanged  by  the  rotation  (II). 

Ans.    x2  +  2/2  +  F  =  0. 

8.  Show  that  no  straight  lines  are  invariant  under  the  rotation  (II). 
Hint.   See  the  hint,  problem  6,  and  apply  Theorem  IV,  p.  90. 

9.  Prove  analytically  that  no  points  are  unchanged  by  a  translation  unless 
all  points  are  unchanged. 

116.  Displacements.  A  transformation  which  replaces  any  figure  by  one 
congruent  to  it  (p.  281)  is  called  a  displacement.  Hence  a  figure  is  displaced 
when  it  is  moved  in  the  plane  from  one  position  to  another.  This  may  evi- 
dently be  accomplished  in  many  different  ways.  Two  displacements  which 
move  a  figure  from  one  position  to  the  same  second  position  are  said  to  be 
equivalent. 

Lemma  I.  A  displacement  is  equivalent  to  a  translation  or  to  a  rotation 
followed  by  a  translation. 

Proof.  Let  the  given  displacement  replace  any  figure  F  by  a  figure  F\ 
Then  if  corresponding  lines  in  F  and  F^  are  parallel  and  have  the  same  direc- 
tion, F  may  be  translated  into  F',  and  hence  the  displacement  is  equivalent 
to  a  translation. 

If  this  is  not  the  case,  then  F  may  be  rotated  into  a  position  F''  such  that 
corresponding  lines  in  F''  and  F^  are  parallel  and  have  the  same  direction 
and  then  F''  may  be  translated  into  F'.  Hence  the  given  displacement  -is 
equivalent  to  a  rotation  followed  by  a  translation.  q.e.d. 


EUCLIDEAN  TRANSFORMATIONS  285 

Theorem  HI.    The  equations  of  any  displacement  have  the  form 
ajf  =  05  cos  ^  —  2/  sin  ^  +  /«,, 


Jaj'  =iccosd  —  y  sind  -\- 
\y^  =z  gc  sin  9  -{■  y  cos  6  + 


(III)  1  .w  _  ^  „;«  fl    I    .,  COS  ^  +  A; 

ly^ere  6,  h,  and  k  are  arbitrary  constants. 

Proof.  Let  the  given  displacement  replace  any  figure  F  by  a  congruent 
figure  F\  Then  by  Lemma  I  it  is  equivalent  to  a  translation  whose  equa- 
tions have  the  form  (III)  when  ^  =  0  (Theorem  I,  p.  282),  or  to  a  rotation 
which  replaces  F  by  a  figure  F''  followed  by  a  translation  which  replaces  F'' 
hy  F\ 

By  Theorem  II, 

x''  =  xcosd  —  y  sin  d,     y''  =  xsmd  +  y  cos  ^, 
and  by  Theorem  I,  x'  =  x"  +  h,     y'  —  y"  +  A:. 

Substituting  the  values  of  x"  and  y"  in  these  equations,  we  obtain  (III). 

Q.E.D. 

If  a  point  is  unchanged  by  a  transformation,  it  is  called  a  fixed  or  an 
invariant  point.     Thus  the  center  of  a  rotation  is  an  invariant  point. 

Theorem  IV.  If  a  displacement  is  not  equivalent  to  a  translation,  there  is  one 
fixed  point. 

Proof  The  point  (x,  y)  will  be  a  fixed  point  when  and  only  when  x'  =  x 
and  y'  =  y.     Substituting  in  (III)  and  transposing,  we  get 

J  (1  —  cos  ^)  X  +  sin  ^  ■  ?/ =  ^, 
'  ^  t  -  sin  ^  •  X  +'  (1  -  cos  d)y=h 

These  equations  can  be  solved,  in  general,  for  one  pair  of  values  of  x  and  y 
(Theorem  IV,  p.  90),  and  hence  there  will  be,  in  general,  but  one  fixed  point. 

-„  ^  ..  1  —  cos  0         sin  d 

But  if  = , 

—  sin  ^        1  —  cos  d 

or,  reducing,  cos  ^  =  1 , 

there  will  be  no  solution,  that  is,  there  is  no  fixed  point.     If  cos  0  =  1,  then 
sin  ^  =  0  (by  3,  p.  19)  and  equations  (III)  become 

x'  =  x-\-h,     y'  =  y-^k, 
which  are  the  equations  of  a  translation. 

Hence  there  is  one  fixed  point  unless  the  displacement  is  a  translation. 

Q.E.D. 

There  cannot  be  an  infinite  number  of  solutions  of  (1)  unless  h  =  k  =  0.    For  if 

1  -  cos  9         sin  0         h 
-  sin  0   "  1  -  cos  e~  Ic' 

then,  as  above,  cos  0  =  1  and  sin  0  =  0.    Substituting  in  (1),  we  get  A  =  0  and  k  =  0.    In  this 
case  every  point  (x,  y)  is  a  fixed  point,  that  is,  there  is  no  displacement. 


286  ANALYTIC  GEOMETRY 

Theorem  V.  Every  displacement  which  is  not  equivalent  to  a  translation  is 
equivalent  to  a  rotation. 

Proof.  If  the  displacement  is  not  equivalent  to  a  translation,  then  it  has  a 
fixed  point  (Theorem  IV).  Let  the  fixed  point  be  chosen  as  origin.  Then  if 
ic  =  0  and  y  ==  0,  we  get  x'  =  0  and  y'  —  0.     Substituting  in  (III),  we  obtain 

^  =  0,     k  =  Q 
as  the  conditions  that  the  origin  is  the  fixed  point.     For  these  values  of  h 
and  k  equations  (III)  reduce  to  (II),  p.  283,  and  hence  the  displacement  is 
equivalent  to  a  rotation.  q.e.d. 

Corollary  I.  Any  two  congruent  figures  may  he  brought  into  coincidence  by 
a  rotation  or  a  translation. 

Corollary  n.  The  perpendicular  bisectors  of  the  lines  joining  corresponding 
points  of  two  congruent  figures  pass  through  the  same  point  or  are  parallel. 

For  if  the  figures  may  be  brought  into  coincidence  by  a  rotation,  they  pass  through  the 
center  of  tlie  rotation;  and  if  the  figures  may  be  brouglit  into  coincidence  by  a  translation, 
they  are  perpendicular  to  the  direction  of  the  translation.  -,  ^ 

PROBLEMS 

1 .  Show  analytically  that  the  angle  between  two  lines  is  unchanged  by  a 
displacement. 

Hint.  Show  that  the  value  of  tan  Q  given  by  (X),  p.  109,  is  an  absolute  invariant  of  the 
displacement  (III). 

2.  Show  analytically  that  the  distance  between  two  points  is  unchanged 
by  a  displacement. 

Hint.   Show  that  the  value  of  I  given  by  (IV),  p.  31,  is  an  absolute  invariant  of  (III). 

3.  Prove  Corollary  II  geometrically  and  derive  Theorem  V  from  it. 

4.  Show  that  a  rotation  about  the  origin  through  an  angle  of  7t  replaces 
any  figure  by  the  figure  symmetrical  to  it  with  respect  to  the  origin. 

5.  Find  the  equations  of  a  rotation  about  the  point  (1,  4)  through  an 

angle  of  -.    Ans.  x'  =  i  VSx -iy +  3  -  i  Vs,  ?/' =  ix  + i  V3y  +  | -2  V3. 
6 

6.  Find  the  equations  of  a  rotation  about  the  point  (3,  —  2)  through  an 
angle  of Ans.   x'  =  y  +  6,  y'  =  —  x  ~\- 1. 

7.  Find  the  equations  of  a  rotation  about  the  point  (xi,  2/1)  through  an 
angle  6.  .  Ans.    x'  =  (x  —  Xi)  cos  6  —  {y  —  y\)  sin  ^  +  Xi, 

y'  =  {x-  Xi)  sin  d  +  {y  -  yi)  cos  d  +  ?/i. 


EUCLIDEAN  TRANSFORMATIONS 


287 


117.  The  reflection  in  a  line.  A  transformation  which  replaces  any  figure 
by  one  symmetrical  to  it  (p.  281)  is  called  a  symmetry  transformation.  The 
simplest  symmetry  transformation  is  the  reflection  in  a  line,  which  replaces 
a  point  by  the  point  symmetrical  to  it  with  respect  to  that  line.  Hence  a 
reflection  in  a  line  replaces  a  figure  by  the  figure  which  is  symmetrical  to  it 
with  respect  to  that  line. 

Theorem  VI.    The  equations  of  a  reflection  in     ^^ 
the  X-axis  are 


(VI) 


W  =  -y' 


a 


C)- 


118,  Symmetry  transformations. 

Lemma  II.  A  symmetry  transformation  is 
equivalent  to  a  reflection  in  any  line  followed 
by  a  displacement. 

Proof.  Let  the  given  transformation  replace 
a  figure  F  by  a  symmetrical  figure  F\    Let  F  be 

transformed  into  a  figure  F''  by  a  reflection  in  any  line.     Then  since  F'  and 
F'^  are  both  symmetrical  to  F,  they  are  congruent  to  each  other. 

For  the  parts  of  F^  and  F^^  are  equal,  since  they  are  equal  to  the  parts  of  F,  and  they 
are  arranged  in  the  same  order,  for  they  are  in  each  case  arranged  in  the  opposite  order 
to  those  of  F. 

Hence  F''  can  be  brought  into  coincidence  with  F'  by  a  displacement, 
that  is,  F  may  be  transformed  into  F'  by  a  reflection  in  any  line  followed 
by  a  displacement.  q.e.d. 


(VII) 


Theorem  VII.    The  equations  of  any  symmetry  transformation  have  the  form 

(oc'  =aocosd  -\-  y  sinO  +  h, 
\y'  =  00  sin6  —  y  cosd  -{-  k, 

where  6,  h,  and  k  are  arbitrary  constants. 

Proof  Let  the  given  transformation  replace  any  figure  i^  by  a  symmetrical 
figure  F'.  Then  by  Lemma  II  it  is  equivalent  to  a  reflection  in  the  JT-axis 
which  replaces  F  by  a  figure  F'%  followed  by  a  displacement  which  replaces 
F''  by  F\ 


By  (VI),  X- 

and  by  (III),  p.  285, 

x'  =  x"  cos  d  —  y''  sin 


y 


h,     y'  =  x''  sin  ^  +  y"  q,q>%Q  ■\-  k. 


Substituting  the  values  of  x''  and  y"  in  these  equations,  we  get  (VII). 

Q.E.D. 


288  ANALYTIC  GEOMETRY 

Theorem  VIII.    The  line  whose  equation  is 

X  cos  w  4-  2/  sin  w  —  p  =  0 

is  transformed  by  (VII)  into  the  line  whose  equation  is 

X  cos  {6  —  uj)  -\-  y  sin  {6  —  w)—  [p  -\-  h  cos {6  —  ca)  +  k  sin  {d  —  w)]  =  0. 

This  is  proved  by  solving  (VII)  for  x  and  y,  substituting  in  the  given  equation,  simpli- 
fying by  9  and  11,  p.  20,  and  dropping  primes. 

A  line  is  said  to  be  invariant  under  a  transformation  if  it  is  transformed 
into  itself  by  that  transformation. 

Theorem  IX.    There  is  always  one  line  which  is  invariant  under  the  sym- 
metry transformation  (VII),  and  if 

h cos le  -\-ksmld  =  0, 
then  all  of  the  lines  perpendicular  to  that  line  are  invariant. 

Proof.    If  the  lines  in  Theorem  VIII  coincide,  then  (Theorem  III,  p.  88) 
cos  w       _      sin  w       _  p 

cos  (d  —  w)      sin  {d  —  u})      p  -\-  h  cos  {d  —  w)  -\-  k  sin  {6  —  w) 

From  the  first  two  ratios 

sin  (6  —  w)  cos  w  —  cos  {9  —  w)  sin  w  =  0, 
or  (9,  p.  20)  sin  (6'  -  2  w)  =  0. 

Hence  ^  —  2  w  =:  0  or  tt, 

.:  (a  =  Id  or  u  =  ^6  —  ^ it. 

Case  I.    ia  —  \d  —  \Tt.     Substituting  this  value  of  w  in  the  last  two  ratios 
of  (1)  and  simplifying  by  4,  p.  19,  and  6,  p.  20,  we  get 
—  cos \Q  _  p 

cos  \Q        p  —  h sin  \d  ■\-  k cos \ 6 
Solving  for  p,  p  =  |  (Ji sin  \d  —  k  cos  i 6). 

Hence  there  is  always  one  pair  of  values  of  w  and  p  for  which  (1)  is  true, 
that  is,  there  is  always  one  line  which  is  transformed  into  itself  by  (VII). 

Case  II.    w  =  i  6.     Susbtituting  this  value  of  w  in  the  last  two  ratios  in 

(1),  we  get  sinig p 

sin  1  ^     p  -{-  h  cos  \d  -\-  k^\n\d 

The  first  of  these  ratios  equals  1,  but  the  second  is  never  equal  to  1  unless 
(2)  hcoQ\e  -\-ksm\e  =  0, 

in  which  case  p  may  have  any  value.  Hence  there  is,  in  general,  but  one 
invariant  line.  But  if  (2)  is  satisfied,  all  of  the  lines  of  a  system  of  parallel 
lines  are  invariant. 

Since  the  values  of  w  in  Case  I  and  Case  II  differ  by  — ,  the  invariant  system 

of  parallel  lines  is  perpendicular  to  the  single  invariant  line.  q.e.d. 


EUCLIDEAN  TRANSFORMATIONS 


289 


Theorem  X.   If  the  invariant  line  of  a  symmetry  transformation  is  the  X-axis^ 

then  the  equations  of  the  transformation  are 


(X) 


(xf  =  cc  -{■  h, 

\yl  =-y. 


Proof.  If  the  X-axis  is  invariant,  then,  if  y  =  0,  we  must  have  y'  =  0  for 
all  values  of  x.  Substituting  y  =  0  and  y'  =  0  in  the  second  of  equations 
(VII),  we  get  X  sin  0  +  A:  =  0. 

This  is  true  for  all  values  of  x  when  and  only  when  sin  ^  =  0  and  A:  =  0. 
If  sin  ^  —  0,  then  cos  ^  =  ±  1- 

Substituting  A;  =  0,  sin  0  =  0,  and  cos  ^  =  1  in  (VII),  we  get  (X). 

Substituting  A:  =  0,  sin  ^  =  0,  and  cos  ^  =  —  1  in  (VII),  we  get 

x'  =  —  x-\-h,    y'  =  y. 

This  transformation  leaves  all  of  the  lines  parallel  to  the  JT-axis  invariant, 
for  if  y  =  a,  then  y'  =  a.  Hence  the  X-axis  is  not  the  single  invariant  line, 
so  that  this  case  is  to  be  excluded ;  that  is,  equations  (VII)  reduce  to  (X)  if  the 
X-axis  is  the  invariant  line  in  Case  I  of  Theorem  IX.  q.e.d. 

Corollary  I.  A  symmetry  transformation  is  equivalent  to  a  reflection  in  a  line 
or  to  a  reflection  in  a  line  followed  by  a  translation  parallel  to  it. 

For  if  ^  =  0,  equations  (X)  reduce  to  equations  (VI). 

If  h  ^  0,  equations  (X)  are  equivalent  to  the  two  transformations 

f  x"  =  x,  -,    x'  =  x"  +  ^, 

■i.     ,,  and 

\y"=-y,  y'=y'\ 

which  are  respectively  a  reflection  in  the  X-axis  and  a  translation  parallel  to  it. 

Corollary  11.  The  middle  points  of  the  lines  joining  corresponding  points  of 
two  symmetrical  figures  lie  on  a  straight  line. 


T  ,'-' 


For  let  (X)  be  the  equations  of  the  symmetry  transformation  which  transforms  one 
figure  into  the  other.    The  middle  point  of  the  line  PP'  is  (Corollary,  p.  39) 
iiix  +  x'),    i(y  +  y')]. 

Substituting  the  values  of  «'  and  y^trom  (X),  this  becomes  (x  +  ^k,  0),  which  is  a  point 
on  the  JC-axis. 


Ix 


290  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Find  the  equations  of  the  curves  symmetrical  to  the  following  curves 
with  respect  to  the  X-axis  and  construct  the  figure.* 

(a)  2/2  -  4  ic  =  0.  (c)  x2  +  4  2/2  -  4  X  =  0. 

(b)  x2  +  0:2/  -  2  2/2  =  0.  (d)  x3  -  8  2/  =  0. 

2.  Show  analytically  that  the  distance  between  two  points  is  unchanged 
by  (a)  a  reflection  in  a  line,  (b)  any  symmetry  transformation. 

3.  Show  analytically  that  the  numerical  value  of  the  angle  which  one 
line  makes  with  another  is  unchanged  by  (a)  a  reflection  in  a  line,  (b)  any 
symmetry  transformation,  but  that  its  sign  is  changed  in  both  cases. 

4.  Find  the  equations  of  the  invariant  lines  which  are  proved  to  exist  in 
Theorem  IX. 

1  9 

5.  Find  the  equations  of  a  reflection  in  the  Y-axis. 

6.  Prove  that  a  reflection  in  a  line  followed  by  a  reflection  in  a  line  per- 
pendicular to  the  first  is  equivalent  to  a  rotation  through  it. 

7.  Asymmetry  transformation  (VII)  has,  in  general,  no  fixed  points,  but 
if  ^(1  4-  cos  ^)  +  jfcsin  ^  =  0,  then  all  of  the  points  of  the  line  x(l—  cos  B) 
—  y  sin  d  =  k  are  fixed  points. 

8.  If  /i  (1  +  cos  e)  -[-Icmid  =  0,  then  (VII)  is  a  reflection  in  a  line. 

9.  Find  the  equations  of  a  reflection  in  the  line  3x  +  42/  —  10  =  0. 

Ans.   X'  =i^x-  §4  2/  +  V,  y'  =  -  ||x  -  J~,y  +  ^f. 

Hint.  The  distances  from  the  line  to  P{x,  y)  and  P'{x\  y^)  (Rule,  p.  106)  must  be 
equal  numerically  with  opposite  signs,  and  the  slope  of  PP'  (Theorem  V,  p.  35)  must  be 
equal  to  the  negative  reciprocal  of  the  slope  of  the  given  line  (Theorem  VI,  p.  36).  These 
conditions  give  two  equations  which  may  be  solved  for  x'  and  y'  in  terms  of  x  and  y. 

10.  Find  the  equations  of  a  reflection  in  the  line  5x  —  12  2/  —  27  =  0. 

Ans.  X'  =  iifx  +  i||2/  +  iih  V'  =  Hf  «  -  my  -  fu- 
ll. Find  the  equations  of  a  reflection  in  the  line  Ax  +  By  -{-  C  =  0. 

,      &-A^  2AB  2AC 

Ans.   X  =  —- —  ^ 


A^+  B^        A'^  +  B^        A^-h  B^ 

2AB  m-A^  2BC 

y  =z  — X y • 

^  ^2  +  2^2        ^2  +  ^^      A^-\-B^ 


EUCLIDEAN  TRANSFORMATIONS 


291 


119.  Congruent  and  symmetrical  conies.  The  conditions  that  two  conies 
should  be  equal  are  given  in  Theorem  XII,  p.  279.     We  shall  now  prove 

Theorem  XI.    Two  equal  conies  are  both  congruent  and  symmetrical. 

Proof.  Since  a  conic  is  symmetrical  with  respect  to  its  principal  axis 
(p.  174),  it  is  unchanged  by  a  reflection  in  that  axis. 

Let  C  and  C  be  two  congruent  conies, 
and  let  D  be  the  displacement  which 
transforms  C  into  C.  Then  C  may  be 
transformed  into  C  by  a  reflection  in 
its  principal  axis  followed  by  the  dis- 
placement D,  that  is  (Lemma  II,  p.  287), 
by  a  symmetry  transformation.  Hence 
C  and  C  are  also  symmetrical. 

Conversely,  let  C  and  C  be  two  sym- 
metrical conies,  and  let  S  be  the  symmetry  transformation  which  transforms 
C  into  C\  Then  S  is  equivalent  to  a  reflection  in  the  principal  axis  of  C 
followed  by  a  displacement  D.  Since  C  is  unchanged  by  a  reflection  in  its 
principal  axis  it  may  be  transformed  into  C"  by  the  displacement  D,  and 
hence  C  and  C"  are  congruent. 

Hence  two  equal  conies  are  both  congruent  and  symmetrical.  q.e.d. 

In  the  figure  C  may  be  transformed  into  C  by  a  rotation  about  0  or  by 
a  symmetry  transformation  consisting  of  (Corollary  I,  p.  289)  a  reflection  in 
the  line  S  which  replaces  C  by  C,  followed  by  a  translation  parallel  to  S. 

120.  nomothetic  transformations.  Given  a  fixed  point  O,  the  transfor- 
mation which  replaces  a  point  P  by  a  point  P'  on  the  line  OP  such  that 

0P'=\-  OP, 


.--'-"A 


0  ^-^, 


::^9^'-^^ 


X<0 


where  X  is  constant,  is  called  a  homothetic  transformation.  0  is  called  the 
center  and  X  the  ratio  of  the  transformation.  Corresponding  figures  are 
called  homothetic  figures.  They  may  easily  be  proved  similar,  with  the  ratio  of 
similitude  (that  is,  the  ratio  of  corresponding  lines)  equal  to  X.  Homothetic 
figures  are  also  similarly  placed. 


292  ANALYTIC  GEOMETRY 

Theorem  xn.    The  equations  of  a  homothetic  transformation  whose  center 
is  {h,  k)  and  whose  ratio  is  \  are 


Proof.    Let  P  and  P'  be  two  corresponding  points.     Then,  by  definition, 
^>.  >  r  OP'=\-OP. 

Projecting  on  the  X-axis  (Theorem  III, 
p.  31), 

x'  —  h  =  \{x  —  li). 
>        Hence         x'  =  'Kx  +  h{l—  X). 

Similarly,  y'  =  \y  -\-k  (1— X).  q.e.d. 


X' 


Corollary.  The  equations  of  a  homothetic  transformation  whose  center  is  the 
origin  and  whose  ratio  is  X  are 

foe'  =Ax,  \2 

\y'  =  Ay. 

121.  Similitude  transformations.  A  transformation  which  replaces  any 
figure  by  one  similar  to  it  is  called  a  similitude  transformation.  It  is  said  to 
be  direct  or  inverse  according  as  corresponding  figures  are  directly  or  inversely 
similar,  that  is,  according  as  the  corresponding  parts  of  the  similar  figures 
are  in  the  same  or  opposite  order. 

If  F  and  F'  are  two  similar  figures  whose  ratio  of  similitude  is  X,  then 
a  homothetic  transformation  with  any  center  and  with  the  ratio  X  will  transform 
F  into  a  figure  F''  which  is  equal  to  F'.  F"  may  be  transformed  into  F' 
either  by  a  displacement  or  by  a  symmetry  transformation  according  as  F' 
and  F"  are  congruent  or  symmetrical,  that  is,  according  as  F  and  F'  are 
directly  or  inversely  similar.     Hence 

Theorem  XIII.  A  similitude  transformation  is  equivalent  to  a  homothetic 
transformation  with  any  center  and  with  its  ratio  equal  to  the  ratio  of  simili- 
tude of  corresponding  figures,  followed  by  a  displacement  or  a  symmetry 
transformation  according  as  the  similarity  is  direct  or  inverse. 

PROBLEMS 

Problems  1  to  4  and  5  to  10  are  to  be  solved  in  order  by  using  those  preceding. 

1.  The  equations  of  a  transformation  of  direct  similitude  have  the  form 

x'  =  X  (x  cos6  —  y  sin  d  +  h),  y'  =  \  {x  sind  -\-y  cos d  +  k). 

2.  A  transformation  of  direct  similitude  has  one  fixed  point. 


I 


EUCLIDEAN  TRANSFORMATIONS  293 

3.  If  the  fixed  point  is  the  origin,  the  equations  of  a  transformation  of 
direct  similitude  have  the  form 

x'=\(xcosd  —  y  sin  6),  y' =\{xsme  +  y  cos  d). 

4.  A  transformation  of  direct  similitude  is  equivalent  to  a  rotation  fol- 
lowed by  a  homothetic  transformation  with  the  same  center. 

5.  The  equations  of  a  transformation  of  inverse  similitude  have  the  form 
^     x'  =\{x cos  6  +  ysind  +  h),  y'  =  \ {x sind  —  y  cos d  +  k). 

6.  The  line  a;  cos  w  +  y  sin  w  —  p  =  0  is  transformed  by  a  transformation 
of  inverse  similitude  into  the  line 

X cos {d  —  <a)  +  y  sin  {d  —  (a)  —  \[p  +  h cos {6  —  cj)  +  k sin {d  —  w)]  =  0. 

7.  The  perpendicular  lines 

{l  —  \)xcos^e  +  {l  —  \)ysinid-\{hcosid  +  k8m^d)=0 
and  {l-\-\)xsmie  -  {1  +  \)ycosi0  -\{hsin^e  —  kcosid)  =0 
are  invariant  under  a  transformation  of  inverse  similitude, 

8.  A  transformation  of  inverse  similitude  has  a  fixed  point. 

9.  If  the  invariant  lines  are  the  axes,  the  equations  of  a  transformation 
of  inverse  similitude  have  the  form  x'  =  \x,  y'  =  —  \y. 

10.  A  transformation  of  inverse  similitude  is  equivalent  to  a  reflection  in 
a  line  followed  by  a  homothetic  transformation  whose  center  is  on  that  line. 

11.  The  equations  of  two  congruent,  symmetrical,  or  similar  curves  are  of 
the  same  degree. 

12.  Show  that  the  angle  which  one  line  makes  with  another  is  unchanged 
by  a  homothetic  transformation. 

13.  Show  that  the  distance  between  two  points  is  multiplied  by  X  by  a 
homothetic  transformation. 

14.  Show  by  means  of  problems  12  and  13  that  a  homothetic  transforma- 
tion is  a  similitude  transformation. 

15.  Show  that  the  angle  which  one  line  makes  with  another  is  unchanged 
by  a  transformation  of  direct  similitude,  but  that  its  sign  is  changed  by  a 
transformation  of  inverse  similitude. 

122.  Similar  conies.  We  have  seen  (Theorem  XI,  p.  291)  that  it  is  un- 
necessary to  distinguish  congruent  and  symmetrical  conies,  and  hence  it  is 
unnecessary  to  distinguish  directly  and  inversely  similar  conies. 


294  ANALYTIC  GEOMETRY 

Theorem  XIV.    If  the  non-degenerate  conic 

Ax^  +  Bxy  +  Cy^-hDx+Ey  -{■F=0 
is  svbjected  to  a  homothetic  transformation  whose  center  is  the  origin  and 
whose  ratio  is  X,  then  the  equation  of  the  homothetic  conic  is 
Ax'^  +  Bxy  +  (72/2  +  XDx  +  \Ey  +  T^^F  =  0. 

This  is  proved  by  solving  the  equations  of  the  transformation  (Corollary,  p.  292)  for  a? 
and  y,  substituting  in  the  given  equation,  and  simplifying. 

Theorem  XV.  If  two  conies  C  and  C  are  homothetic^  the  origin  being  the 
center  and  X  the  ratio,  then 

H'2  H^  H2  H3 

where  —  and  —  are  the  absolute  invariants  of  C\  and  —  and  —  are  those 

A'  0'  "^       '  A  0 

of  C* 

Proof.    Let  the  equation  of  C  be 

Ax'^  +  Bxy  +  Cy'^  +  I)x  +  Ey  +  F  =  0, 
and  then  by  Theorem  XIV  that  of  C  may  be  written  in  the  form 
^x2  +  Bxy  +  C?/2  +  XDx  +  \Ey  +  \^F  =  0. 
The  absolute  invariants  of  C  are 
H^2  _   (^  +  C)2   _  H2 
A'  ~  m-^AC~  A  ' 

H^ {A  +  Cf -15! 

Q'  ~  AA  CX^F  +  BXDXE  -  A  (XE)^  -  C  (XDf  -  X^FB^  ~  X'^  0  ' 

Q.E.D. 

Theorem  XVI.  If  two  non-degenerate  conies  C  and  C  are  similar^  then  their 
absolute  invariants  and  their  ratio  of  similitude  X  satisfy  equations  (XV), 
Conversely,  if  the  absolute  invariants  of  two  conies  C  and  C  satisfy  the  first  of 
equations  (XV)  and  if  the  value  of  X  determined  by  the  second  is  real,  then  C 
and  C  are  similar,  with  the  ratio  X. 

Proof.  By  Theorem  XIII,  p.  292,  C  may  be  transformed  into  C  by  a 
homothetic  transformation,  whose  center  is  the  origin  and  whose  ratio  is  X, 
which  transforms  C  into  a  conic  C'\  followed  by  a  displacement  or  symmetry 
transformation  which  transforms  C  into  C.     Then,  by  Theorem  XV, 

^  '  A^~  a'      0^~X2  0' 

and  by  Theorem  XII,  p.  279, 

H-2_Hr2    Er3_ir3 

^'  A''  ~  A'  '     0''  ~  0'  * 

*  Theorem  V,  p.  272.    The  values  of  A,  H,  and  ©  are  given  on  p.  264. 


EUCLIDEAN  TKANSFOKMATIONS  295 

From  equations  (1)  and  (2)  we  obtain  equations  (XV). 

Conversely,  if  equations  (XV)  are  satisfied,  the  value  of  X  determined  by 
the  second  being  real,  then  C  and  C  are  similar.  For  let  C  be  transformed 
into  a  conic  C"  by  a  homothetic  transformation  whose  center  is  the  origin 
and  whose  ratio  is  \.*  Then  equations  (1)  are  true  by  Theorem  XV.  From 
(1)  and  (XV)  we  get  equations  (2),  and  hence  (Theorem  XII,  p.  279)  C  and 
C  are  equal.  Then  C  may  be  transformed  into  C  by  either  a  displacement 
or  a  symmetry  transformation.  Hence  C  may  be  transformed  into  C  by  a 
homothetic  transformation  followed  by  a  displacement  or  a  symmetry  trans- 
formation, that  is  (Theorem  XIII,  p.  292),  by  a  similitude  transformation. 
Hence  C  and  C  are  similar.  q.e.d. 

Corollary  I.  Two  conies  are  similar  if  the  coefficients  of  the  terms  of  the 
second  degree  are  proportional,  that  is,  if 

and  the  value  of  X  determined  by  the  second  of  equations  (XV)  is  real. 
For  if  r  is  the  common  value  of  these  ratios,  then 

A  =  rA',    B^rB',    C  =  rC', 

H2  (rA'  +  rCQ^ r^jA'+Cy         W^ 

and  hence  ^  -  ^^^^^^  _  4  rA'rC  ~  r^  {B'^  -  4  A'C)  ~  A'  ' 

Hence  the  first  of  equations  (XV)  is  satisfied. 
Corollary  II.    Any  two  parabolas  are  similar. 

For  if  C  and  C  are  parabolas,  then  (Theorem  IX,  p.  277)  A  =  0  and  A'  =  0.  Hence  the 
first  of  equations  (XV)  is  satisfied.  Since  A  =  0,  H  and  ©  have  opposite  signs  (p.  279)  and 
similarly  H'  and  0'  have  opposite  signs.  Hence  the  value  of  A.  obtained  from  the  second 
of  equations  (XV)  is  real. 

Ex.  1.  Show  that  the  conies  x^+2y^=36  and  3x'2+2xy-\-3y^-6x-2y-5=0 
are  similar  and  find  the  ratio  of  similitude. 

Solution.  Computing  the  absolute  invariants  of  the  given  equations  and  substi- 
tuting in  (XV),  we  obtain 

_(6)2__j3)2        (6)3    ^  1      (3)8 
-  32  ~  -  8 '     -  256  ~  X2  -  288 ' 

Solving  the  second  equation,  we  get  X=  ±  ^.  Hence  the  first  of  equations  (XV) 
is  satisfied,  and  the  second  is  satisfied  if  X  =  ±  i.  The  conies  are  therefore  similar, 
with  the  ratio  of  similitude  equal  to  ±  i.  The  double  sign  means  that  they  are 
either  directly  or  inversely  similar. 

*  The  proof  would  break  down  at  this  point  if  the  value  of  \  determined  by  the  second 
of  equations  (XV)  were  imaginary,  because  the  ratio  of  a  homothetic  transformation  is 
a  real  number. 

That  such  cases  arise  is  illustrated  by  the  hyperbolas  ix^-  ]p  =  16  and  —  4  a:^  + 1/2  =  4, 
whose  absolute  invariants  satisfy  the  first  of  equations  (XV) ;  but  from  the  second, 


296  ANALYTIC  GEOMETRY 

PROBLEJMS 

1.  Show  that  the  following  pairs  of  conies  are  similar.     Find  the  ratio 
of  similitude  in  each  case  and  construct  the  figure. 

(a)  x^-4y^  =  l,  3  x2  +  4  x?/  +  4  =  0. 

(b)  x2  +  4 y  ^  0,  y^  -8x  =  0. 

(c)  9x2  4-2/2^9^  x2  +  92/2  _  54  7/  =  0. 

(d)  16  x2  +  9  2/2  =  144,  25  x2  +  14  xy  +  25  ?/2  =  72. 

(e)  x2  -  2/2  =  a2,  2  xy  =  a'2. 


^?2S. 

X=±2.- 

Ans. 

X  =  ±2. 

Ans. 

X=±3. 

Ans. 

X=±i. 

Ans. 

Ans. 

^  =  ±? 

i> 


(f)  2/2  =  2px,  (X  -  /i)2  =  2p\y  -  k). 

2.  Show  that  the  ellipses  in  Ex.  1,  p.  200,  are  similar. 

3.  Show  that  the  hyperbolas  in  Ex.  2,  p.  201,  for  which  k  is  positive  or 
for  which  k  is  negative,  are  similar. 

4.  Show  that  the  locus  of  Ax^  +  Bxy  +  Cy^  =  k  is,  in  general,  a  system 
of  similar  conies.  Discuss  all  possible  special  cases  in  which  this  statement 
is  not  exact. 

5.  Any  homothetic  transformation  is  equivalent  to  a  homothetie  trans- 
formation whose  center  is  the  origin  followed  by  a  translation. 

6.  By  means  of  problem  4  prove  that  two  conies  are  homothetic  if  the 
coefficients  of  the  terms  of  the  second  degree  are  proportional. 

7.  Find  the  center  and  ratio  of  the  homothetic  transformation  which 
transforms  y^  =  2px  into  2/^  =  2_p'x.  ^^^^    /q  q\    x  =  :^, 

8.  A  homothetic  transformation  whose  center  is  0(0,  0)  and  whose  ratio 
is  X  followed  by  a  homothetic  transformation  whose  center  is  0'{a,  0)  and 
whose  ratio  is  X'  is  equivalent  to  a  homothetic  transformation  whose  center 

is  ( »  0  I ,  that  is,  a  point  on  00'  and  whose  ratio  is  XX'. 

V 1  -  XX'      / 

9.  A  circle  may  be  transformed  into  any  other  circle  by  two  homothetic 
transformations  whose  centers,  called  the  centers  of  similitude  of  the  circles, 
lie  on  the  line  of  centers. 

Hint.  Take  the  center  of  one  circle  for  the  origin  and  let  the  JT-axis  pass  through  the 
center  of  the  other  circle.  Substitute  from  (XII),  p.  292,  in  the  equation  of  the  first 
circle  and  determine  h,  k,  and  A  so  that  the  result  coincides  with  the  second  circle. 

10.  Given  three  circles,  the  line  joining  a  center  of  similitude  of  one  pair 
with  a  center  of  similitude  of  a  second  pair  will  pass  through  a  center  of 
similitude  of  the  third  pair. 

Hint.   Apply  problem  8. 

1 1 .  The  six  centers  of  similitude  of  three  circles  taken  by  pairs  lie  three 
by  three  on  four  straight  lines. 

Hint.   Apply  problem  10. 


CHAPTER  XIV 


INVERSION 


123.  Definition.   Let  0  be  a  given  point  and  let  P  be  any  point  of  a 
figure  F.     Construct  P'  on  OP  such  that 

OP'  •  OP  =  1. 

By  letting  P  assume  different  positions  on  P, 

P"  will  move  on  a  figure  F\     The  operation  or 

transformation  which  replaces  P  by  P'  is  called 

an  inversion,  while  F  and  F'  are  called  inverse 

figures.     0  is  called  the  center  of  the  inversion. 

\  /  The  figure  has  been  accurately  constructed  and  indi- 

^v  ^^"^  cates  tliat  the  inverse  of  a  triangle  is  a  figure  bounded 

"'        '  by  three  curves.    Hence  we  may  expect  to  find  that  the 

properties  of  inverse  figures  are,  in  general,  quite  different  from  those  of  equal  or 

similar  figures. 

Two  important  properties  of  an  inversion  are  immediately  evident  from 
the  definition. 

1.  If  P  approaches  the  origin,  Pi  recedes  to  infinity,  and  conversely. 

For  if  OP  approaches  zero,  then  OP'  must  become  infinite  since  OP'  •  OP  =  1,  and 
conversely. 

2,  The  points  of  the  circle  of  unit  radius  whose  center  is  0  are  fixed  points. 

For  if  OP  =  1,  then  from  OP'  .  OP  =  1  we  get  OP'  =  1.  Hence  P'  coincides  with  P, 
that  is,  P  is  a  fixed  point.  This  fact  is  useful  in  plotting  inverse  figures,  for  the  points 
in  which  a  figure  cuts  this  circle  will  be  points  of  the  inverse  figure. 

1 24.  Equations  of  an  inversion.  By  the  equations  of  an  inversion  we  mean 
two  equations  involving  the  coordinates  of  two  corresponding  points  P  and 
P'.     These  equations  must  express  the  two  conditions  : 

1.  That  P  and  P'  lie  on  a  line  through  the  center. 

2.  That  OP'    0P  =  1. 
The  first  of  these  conditions  is  satisfied  when  the 

triangles  0PM  and  OP'M'  are  similar,  whence 
X  _y  _  OP 
x'~y'~~dP'' 


(1) 


p'(:^',y') 


(2) 


The  second  condition  may  be  written,  by  dividing  by  OP'^ 
OP  _     1     _        1 

OP'"   OP'2"~x'2  +  y'2' 

297 


(by  (IV),  p.  31) 


298 


ANALYTIC  GEOMETKY 


From  (1)  and  (2), 


V  ^        1 

y'      x'2  +  2/'2' 


;:;'  V 


y 


Hence  we  have 

Theorem  I.    The  equations  of  an  inversion  whose  center  is  the  origin  are 


(I) 


_        X'  _        V 


Ex.  1.    Find  the  inverse  of  the  line  2a;  +  4y  —  1  =  0. 

Solution.    Substitute  the  values  of  x  and  y  given  by  (I)  in  the  given  equation. 
We  thus  obtain 


2x' 


+ 


4?/' 


1  =  0. 


Reducing  and  dropping  primes,  we  get 
a;2  +  2/2  _  2x  — 4?/  =  0. 

This  is  the  equation  of  a  circle_  whose  center  is  the 
X    point  (1,  2)  and  whose  radius  is  Vs  (Theorem  I,  p.  131). 
In  the  figure  a  number  of  inverse  points  are  indicated 
by  the  dotted  lines. 

Ex.  2.    Find  the  inverse  of  the  straight  line  Ax  +  By  +  C  —  0. 
Solution.    Substitute  in  the  given  equation  the  values  of  x  and  y  given  by  (I). 
This  gives 


Ax' 


+  C7=0. 


Simplifying  and  dropping  primes, 

Cx^  +  Cy^  +  Ax-\-By  =  0. 

The  locus  of  this  equation  is  a  circle  (Theorem  II,  p.  132)  which  passes  through 
the  origin  (Theorem  VI,  p.  73).    If  C  =  0,  the  locus  is  the  given  line.    Hence 

The  inverse  of  a  straight  line  which  does  not  pass  through  the  origin  is  a  circle, 
and  a  line  which  passes  through  the  origin  is  invariant  under  an  inversion. 


Ex.  3.    Find  the  inverse  of  the  circle  x^  +  y"^  -{-  Dx  -\- 
Solution.    Substituting  from  (I),  we  get 


4- 


Dx' 


W 


+  i^=0. 


+  i^=0. 


(x'2  +  l/'2)2         (a.-2  +  y/2)2         cc'2  +  |/'2         ^"^  ^  y'l 

Multiplying  by  x"^  +  y"^  and  dropping  primes, 
(3)  Fx'^  +  i^?/2  +  Dx  +  JE;?/  +  1  =  0. 

The  locus  is  a  circle  (Theorem  II,  p.  132)  unless  J^  =  0,  in  which  case  (3)  is  an 
equation  of  the  first  degree  and  its  locus  is  a  straight  line  (Theorem  II,  p.  86) .    Hence 

The  inverse  of  a  circle  is,  in  general,  a  circle,  but  the  inverse  of  a  circle  which 
passes  through  the  origin  is  a  straight  line. 


INVERSION 


299 


PROBLEMS 

1.  If  the  origin  is  the  center  of  inversion,  find  the  inverse  of  each  of  the 
following  curves.     Construct  the  figure  in  each  case. 


(a)  2x  =  l. 

(b)  42/ =  1. 

(c)  X  +  y  -  1  =  0. 


(d)  x2  + 

(e)  x2  +  ?/  =  4. 

(f )  x2  +  2/2  -  2  X 


4x  =  0. 


4  2/  +  1  =  0. 


(g)  x2  +  2/2  +  4  X  - 
(h)  3x-42/  =  0. 

(i)  x^-y^  =  0. 

(j)  4x-32/  =  l. 
(k)  x2  +  2/2  +  2  2/ : 

(1)  2/2  =  4  X. 


62/ 


0. 


I 


2.  Find  the  inverse  of  the  points  (0,  2),  (3,  0),  (3,  4),  (2,  1),  (i,  0),  (i,  i), 
(a,  0),  and  (0,  b).     Plot  the  given  and  inverse  points. 

3.  Prove  by  (I)  that  the  points  on  the  unit  circle  are  fixed  points. 

4.  Find  the  equation  of  all  circles  which  are  unchanged  by  an  inversion 
whose  center  is  the  origin.  Ans.    x"^  -\-  y^  +  Dx  -}-  Ey  +  1  =  0. 

5.  Show  that  the  inverse  of  the  center  of  a  circle  is  not,  in  general,  the 
center  of  the  inverse  circle. 

6.  Show  that  the  center  of  the  circle  obtained  in  Ex.  2  lies  on  the  perpen- 
dicular drawn  from  the  origin  to  the  given  line. 

7.  Show  that  the  inverse  of  a  circle  whose  center  is  the  center  of  inver- 
sion is  a  concentric  circle. 

125.  Inversion  of  conic  sections.  In  this  section  we  shall  discuss  several 
curves  which  are  obtained  by  inverting  a  conic  section.  These  curves  have 
been  otherwise  defined  in  Chapter  XI. 

Theorem  II.  The  inverse  of  the  parabola  is  the  cissoid  if  the  vertex  of  the 
parabola  is  the  center  of  inversion. 

Proof.    If  the  vertex  of  the  parabola  is  the  origin,  its  equation  is 

y^  =  2px. 
Then,  from  (I),  p.  298, 


r 


2px' 


(X'2  +  2/'2)2         x'2  +  y'2 

Reducing  and  dropping  primes. 

This  is  the  equation  of  the  cissoid  of  Diodes 

(problem  10,  p.  253).    If  we  replace  —  by  2  a, 

2p 

we  obtain  the  form  of  the  equation  usually 

given,  namely, 

(1)  x^  =  y^{2a-x).  q.e.d. 


300 


ANALYTIC  GEOMETRY 


A  general  discussion  (p.  74)  gives  us  the  following  properties  of  the 
cissoid. 

1.  The  cissoid  passes  through  the  origin  (Theorem  VI,  p.  73). 

2.  It  is  symmetrical  with  respect  to  the  X-axis  (Theorem  V,  p.  73). 

3.  Its  intercepts  on  both  axes  are  zero  (Rule,  p.  73). 

4.  The  cissoid  lies  entirely  between  the  F-axis  and  the  line  x  =  2  a. 
For,  solving  (1)  for  y, 


(2) 


-~^<^ 


If  X  is  negative,  tlie  numerator  is  negative  and  the  denominator  positive ;  and  if  a;  >  2  a, 
the  numerator  is  positive  and  the  denominator  negative.  In  either  case  the  fraction  is 
negative  and  y  is  imaginary. 

5.  The  cissoid  recedes  indefinitely  from  the  X-axis  and  approaches  the 
line  x=  2  a. 

For  as  x  approaches  2  a  the  fraction  in  (2)  becomes  larger  and  approaches  infinity  as  a 
limit. 

This  may  also  be  seen  by  transforming  (1)  to  polar  coordinates,  vrhich  gives 

p=  2a  sine  tan  0 

as  the  polar  equation  of  the  cissoid  ;  and  hence,  if  0  =  -  or  —  >  p  =  oo. 

Theorem  III.     The  inverse  of  the  equilateral  hyperbola  is  the  lemniscate  if 
the  center  of  inversion  is  the  center  of  the  hyperbola. 

Proof.    The  equation  of  the  equilateral  hyperbola  is  (p.  186) 

x2  -  2/2  =  a2. 

The   equation  of  the  inverse 
curve  is  (by  (I),  p.  298) 


Reducing  and  dropping  primes, 


(X2  +  2/2 


(X2-2/2). 


The  locus  is  the  lemniscate  of 
Bernoulli  (problem  1,  (g),  p.  248, 
and  problem  4,  p.  262) .    Replacing 


by  a'2,  we  get  the  form  of  the  equation  usually  given,  namely, 


(3)  (X2  +  2jY  =  a'2  (x2  -  2/2).  Q.E.D. 

A  discussion  of  the  equation  of  the  lemniscate  in  polar  coordinates  is  given  in  Ex.  2, 
p.  152.  From  (3)  it  is  evident  that  the  lemniscate  is  symmetrical  with  respect  to  both  axes 
and  the  origin  (Theorem  V,  p.  73). 

In  the  figure  a<  1  and  a' >  1.  If  a  =  a'=l,  the  lemniscate  will  be  tangent  to  the 
hyperbola  at  its  vertices.    If  a  >  1  and  a'  <  1,  the  two  curves  will  not  intersect. 


INVERSION 


301 


Theorem  IV.    The  inverse  of  the  equilateral  hyperbola  is  the  strophoid  if  the 
center  of  inversion  is  a  vertex  of  the  hyperbola. 

The  equation  of  the  equilateral  hyperbola,  when  the  origin  is  the  right- 
hand  vertex,  is 

x2  -  2/2  +  2  ax  =  0. 

This  is  obtained  from  a;2 -  ?/2  =  a^  by  setting  (Theorem  I,  p.  160)  x-  x'  +  a,y  =  y',  and 
dropping  primes. 


The  inverse  curve,  from  (I),  p.  298,  is 


2nx' 


(X^2  +  y^2y2         (;^^2  +  y^2y2 

Reducing  and  dropping  primes, 

1 

0. 


0. 


X(x2  +  ?/2)  +  -— (X2-2/2 

2  a 


The  locus  of  this  equation  is  the 
strophoid  (problem  9,  p.  262).    Repla- 
cing —  by  a'  and  solving  for  y^,  we 
2a 

get  the  form  of  the  equation  usually 
given,  namely. 


(4) 


y 


.,  a  +  X 

x^ 

a'  —  X 


Q.B.D. 


In  the  figure  a'=2a  =  l.    If  a'  >1  and 
2a  <  1,  the  left-hand  branch  of  the  hyperbola 

will  intersect  the  loop  of  the  strophoid.    If  a'  <  1  and  2  a  >  1,  the  left-hand  branch  of  the 
hyperbola  will  not  meet  the  strophoid. 

A  general  discussion  of  (4)  gives  us  the   following  properties  of  the 
strophoid. 

1.  It  passes  through  the  origin  (Theorem  VI,  p.  73). 

2.  It  is  symmetrical  with  respect  to  the  X-axis. 

3.  Its  intercepts  are'?/  =  0  or  0  and  x  =  —  a',  0,  or  0.     Hence  it  passes 
twice  through  the  origin. 

4.  The  strophoid  lies  entirely  between  the  lines  x  =  a'  and  x  =  —  a\ 

For,  solving  (4)  for  y, 

(5) 


Va'  +  X  X  /—- 
=  ± Va'2  -  x^ . 
a'  —  X         a'  -  X 


The  quadratic  under  the  radical  is  negative  for  values  of  x  not  lying  between  the 
roots  (Theorem  III,  p.  11),  and  for  these  values  y  is  imaginary. 

5.  The  strophoid  recedes  indefinitely  from  the  X-axis  and  approaches  the 
line  X  =  a\ 

For,  from  (5),  y  becomes  infinite  when  x  approaches  a'. 


302 


ANALYTIC  GEOMETRY 


Theorem  V.    The  inverse  of  a  conic  is  a  limaQon  if  the  center  of  inversion  is 
a  focus  of  the  conic. 

Proof.    The  equation  of  a  conic  whose  focus  is  the  origin  is  (Theorem  II, 
p.  178) 

(1  -  e2)  x2  +  2/2  _  2  e^px  -  e^p^  =  0. 

Substituting  from  (I),  p.  298,  the  equation  of  the  inverse  curve  is 


(l-e2)x'2 


2  e'^px' 


-  e2p2  =  0. 


(X'2  +  2/'2)2         (x'2  4.  y^y         a;'2  +  y^2 

Clearing  of  fractions,  transposing,  and  dropping  primes, 

e2p2  (x2  4-  2/2)2  4_  2  e^x  {x^  +  2/2)  =  (1  -  e2)  x^  +  2/2. 
Adding  6^x2  to  both  sides  and  dividing  by  e2_p2^ 

(x2  +  2/2  +  ia:)2  =  J-(x2  +  y2). 
p  e2p2 

The  locus  of  this  equation  is  the  lima9on  (problem  11,  p.  253).     If  we  set 

—  =  a  and  — —  =  b^,  we  get  the  form  of  the  equation  usually  given,  namely, 
p  e2p2 

(6)  (x2  +  2/2  +  ax)2  =  62  (a;2  +  yzy  q  e.d. 

The  lima9on  has  three  distinct  forms  corresponding  to  the  three  forms  of  conies, 
according  as  a  is  less  than,  equal  to,  or  greater  than  b.  If  a=  b,  the  lima9on  is  some- 
times called  the  cardioid  (Ex.  2,  p.  158). 


a  >b 


A  general  discussion  of  (6)  gives  us  the  following  properties  of  the  limagon. 

1.  It  passes  through  the  origin  (Theorem  VI,  p.  73). 

2.  It  is  symmetrical  with  respect  to  the  X-axis  (Theorem  V,  p.  73). 

3.  Its  intercepts  are  x  =  0,  0,  —  a  —  b,  and  —  a  +  6  and  y  =  0,  0,  6,  and 
b.     Hence  the  lima^on  passes  twice  through  the  origin. 

4.  The  lima9on  is  a  closed  curve. 

For,  transforming  to  polar  coordinates,  (6)  becomes  (Theorem  I,  p.  155) 

(p2  +  ap  cos  0)2  =  62p2. 

Solving  for  p,  p  =  6  -  a  cos  fl. 

Since  —  1  ^  cos  0  ^1,  p  cannot  become  infinite. 


INVERSION 


303 


PROBLEMS 

1.  Construct  the  following  conies,  find  the  equations  of  the  inverse  curves, 
and  discuss  and  construct  their  loci. 

(a)  y2  =  a;,  2/2  =  8a;,  x^  =  iy. 

(b)  x^  -  y^  =  i,  x^  -  y^  =  I,  x^  -  ?/  =  l,2xy  =  l. 

(c)  x2  -  2/2  +  V2  x  =  0,  a;2  _  2/2  +  X  =  0,  x2  -  ?/2  +  4  X  =  0. 

(d)  3x2  +  4?/2-4x  =  4,  2/2 -4x  =16,  3x2  -  2/2  +  16x  +  16  =  0. 

2.  Find  the  inverse  of  the  hyperbola  Srx^  —  ry^  +  2x  =  0,  and  discuss  its 


properties. 


Ans.    The  trisectrix  of  Maclaurin  x  (x2  +  2/^)  =  -  (2/'"^ 


3x2). 


3.  Prove  that  the  inverse  of 

(a)  the  cissoid  is  a  parabola  ; 

(b)  the  lemniscate  is  an  equilateral  hyperbola ; 

(c)  the  strophoid  is  an  equilateral  hyperbola  ; 

(d)  the  limagon  is  a  conic,  if  the  origin  is  the  center  of  inversion. 

4.  Prove  analytically  and  geometrically  that  if  a  curve  C  inverts  into  C, 
then  C  inverts  into  C. 

5.  Show  that  the  inverse  of  the  locus  of  an  equation  of  the  second  degree 
is,  in  general,  a  curve  whose  equation  is  of  the  fourth  degree.  In  what 
combination  of  x  and  y  will  the  terms  of  the  fourth  degree  enter  ?  What 
will  be  the  degree  if  the  given  locus  passes  through  the  origin  ? 


126.  Angle  formed  by  two  circles.  If 
radii  be  drawn  to  a  point  of  intersection  of 
two  circles,  the  angle  formed  is  equal  to  one 
of  the  angles  formed  by  the  tangents  at  that 
point,  since  their  sides  are  respectively  per- 
pendicular. That  angle  d  is  called  the  angle 
formed  by  two  circles. 

Theorem  VI.  The  angle  6  formed  hy  two 
intersecting  circles 

Ci :  x2  +  2/2  +  1^1^  +  ^12/  +  Fi  =  0 
and  Oa  :  x2  +  2/2  +  DgX  +  ^22/  +  -^^2  =  0 

is  given  hy 


(VI) 


COS 


e 


J>il>2  +  E^E^  -^F^-^F^ 


Vl>i2  _|_  J5:^2  _  4  p^  ^Jj2  _,_  jg;^2  _  4  j^^ 


304 


ANALYTIC  GEOMETRY 


Proof.    By  definition  d  equals  the  angle  formed  by  tlie  radii  drawn  to  a 
point  of  intersection.     Hence  from  tlie  figure  and  17,  p.  20, 


(1) 


2rir2 


where  ri  and  r^  are  the  radii  of  C\  and  C2  respectively  and  d  is  the  length  of 

the  line  of  centers.     By  Theorem  I,  p.  131, 


r2-iVD22  +  ^22_4j.2, 
and  the  centers  of  C\  and  C%  are  respec- 
«ve„(-f,-f)a„d(-f,-f). 
Hence  (by  (IV),  p.  31) 

Substituting  in  (1)  and  reducing,  we  get  (VI).  ^  q.e.d. 

Corollary.    Ci  and  C2  are  orthogonal  if  BiD^  +  J&i£'2  -  2  i^i  -  2  i^2  =  0. 

127.  Angles  invariant  under  inversion. 

Theorem  VII.    The  angle  between  two  circles  is  equal  to  the  angle  formed  by 
the  inverse  circles. 

Proof.    Let  the  equations  of  two  circles  be 

Oi :  x2  +  ?/2  +  Dix  +  Eiy  +  F^  =  0 
and  C2 :  x^  +  2/2  +  I)2X  +  E2y  +  F2  ==  0. 

Then  the  equations  of  the  inverse  circles  are  respectively  [(3),  p.  298] 

J_ 
Fi 
1 


Ci':.2  +  ,.  +  ^^  +  ^l 


+ 


0 


and 


C2':x2  +  y2  +  :^a; 
F2 


-F2  i'a 


By  Theorem  VI  the  angle  formed  by  Ci  and  C2'  is  given  by 

^  _  2 
Fi       F2 


cos  r  = 


D1D2  .  E\E2 
F1F2       F1F2 


Mhrn-rMhrnH 


D1D2  +  -E^ijEJa  -  2  i?'i  -  2  F2 


VDi2  +  J?i2 

cos  ^. 


4  Fi  VDs^  +  ^2^  -  4  i<^2 


where  6  is  the  angle  formed  by  Ci  and  C2. 

Since  6'  and  ^  are  both  less  than  7t,  we  therefore  have  6'  =  6. 


Q.E.D. 


INVERSION  305 

Corollary.  The  angles  formed  by  two  intersecting  curves  are  equal  to  the 
angles  formed  by  the  inverse  curves. 

For  draw  two  circles  respectively  tangent  to  the  given  curves  at  a  point  of  inter- 
section. The  inverse  circles  will  be  tangent  to  the  inverse  curves  at  a  point  of  intersec- 
tion. The  angles  formed  by  either  pair  of  curves  and  the  tangent  circles  are  identical, 
and  the  angles  formed  by  the  two  pairs  of  circles  are  equal.  Hence  the  angles  formed 
by  the  given  curves  and  by  the  inverse  curves  are  equal. 


PROBLEMS 

1.  Find  the  angles  formed  by  the  following  pairs  of  curves  and  the  angles 
formed  by  the  inverse  curves,  and  show  that  they  are  equal. 

(a)  x-y  =  0,  x  +  2y  =  0. 

(b)  x-\-3y  -2  =  0,  x-2y  =  0. 

(c)  x^  +  y^-^ix-Sy  =  0,  x^-hy^-^x  =  0. 

(d)  x^+y^-4x  +  12  =  0,  x^  +  y^  -8y  =  0. 

(e)  x2  -f  2/2_6x  +  4y  =  0,  6x-4y-l  =  0. 

2.  Show  that  the  circles  found  in  problem  4,  p.  299,  are  orthogonal  to  the 
circle  x^  -{-  y'^  =  1. 

3.  If  P  and  P'  are  two  inverse  points,  show  that  all  of  the  circles  which 
pass  through  P  and  are  orthogonal  to  x^  +  y^  =  1  will  also  pass  through  P\ 

4.  How  may  problem  3  be  used  to  define  an  inversion  ?  * 

5.  Into  what  kind  of  a  figure  will  three  lines  forming  a  triangle  invert  if 
the  center  of  inversion  is  not  on  one  of  these  lines  ? 

6.  Into  what  kind  of  a  figure  will  three  circles  which  have  a  point  in 
common  invert  if  that  point  is  the  center  of  inversion  ? 

7.  Three  circles  pass  through  a  point  and  intersect  each  other  in  three 
other  points.  Show  that  the  sum  of  the  angles  formed  by  the  circles  at 
these  three  points  is  two  right  angles. 

Hint.  Invert  the  figure,  using  the  point  common  to  the  three  circles  as  the  center  of 
inversion. 

8.  Three  circles  pass  through  the  same  point.  Show  how  to  construct 
four  circles  tangent  to  the  three  given  circles. 

Hint.  Suppose  the  required  circles  constructed.  Invert  the  figure,  using  the  common 
point  as  the  center  of  inversion  and  show  how  to  construct  the  inverse  of  the  required 
circles.    Then  invert  the  figure  so  constructed,  using  the  same  center  of  inversion. 

9.  Show  that  the  sign  of  an  angle  is  changed  by  an  inversion. 


306 


ANALYTIC  GEOMETRY 


128.  Inversion  of  systems  of  straight  lines. 

Theorem  "VIII.  The  inverse  of  a  system  of  parallel  lines  is  a  system  of  tan- 
gent circles  whose  centers  lie  on  a  line  perpendicular  to  the  lines  of  the  system. 

Proof  Choose  one  of  the  lines  of  the  system  for  the  Y-axis.  Then  the 
equation  of  the  system  is  x  =  a,  where  a.  is  an  arbitrary  constant.    The 


inverse  system  is  therefore  (by  (I),  p.  298) 


x'2  +  y's 


a,  or,  reducing  and 


X' 


dropping  primes,  cc^  +  ?/2 x  =  0.     This  is  the  equation  of  a  system  of 

circles  whose  centers  lie  on  the  X-axis  and  which  are  tangent  to  each  other 
at  the  origin  (Theorem  VIII,  p.  144).  q.e.d. 

Theorem  IX.  The  inverse  of  a  system  of  lines  passing  through  a  point  is  a 
system,  of  circles  passing  through  the  origin  and  through  the  inverse  of  that 
point. 

Proof  Let  the  system  of  lines  be  y  =  mx  +  6,  where  h  is  constant  and  m 
varies.  By  Theorem  I,  p.  298,  the  inverse  of  the  system  is,  after  reducing 
and  dropping  primes, 


x-^  +  2/2  + 


y  =  0. 


This  is  the  equation  of  a  system  of  circles  passing  through  the  origin 
(Theorem  VI,  p.  73)  and  through  (0,  t  )  (Corollary,  p.  53),  which  is  the 
inverse  of  (0,  b)  through  which  the  lines  pass,  q.e.d. 


INVERSION 


307 


129.  Inversion  of  a  system  of  concentric  circles. 

Theorem  X.  The  inverse  of  a  system  of  concentric  circles  is  a  system  with 
two  limiting  points,  one  at  the  origin  and  the  other  at  the  inverse  of  the  center 
of  the  concentric  circles. 

Proof.    The  equation 

(1)  ic2  -f  2/2  _  2  /3?/  +  (82  -  r2  =  0 

represents  a  system   of  concentric   circles  if  j8  is  constant  and  r  varies 

(Tlieorem  II,  p.  58). 


(2) 


The  inverse  of  (1)  is  [(3),  p.  298] 

2^ 


X'^  +  y-^ 


^2 


y  + 


The  locus  of  (2)  is  a  system  of  circles  with  their  centers  on  the  F-axis 
(Corollary,  p.  131).     The  radius  of  any  one  is  (Theorem  I,  p.  131) 


r" 


2  \\B2-r^/ 


Hence  r'=  0  if  r  =  0,  and  the  locus  of  (2)  is  then  the  point-circle  (0,  -  j » 

which  is  the  inverse  of  (0,  /3),  the  center  of  (1).  If  r  =  co,  (2)  becomes 
x'2  _j_  ^2  —  0,  whose  locus  is  the  origin.  Hence  the  system  (2)  has  two  limiting 
points  (p.  144),  at  the  origin  and  at  the  inverse  of  the  center  of  (1).       q.e.d. 


PROBLEMS 

1.  Why  do  we  not  consider  tlie  system  of  lines  passing  through  the  origin 
in  proving  Theorem  IX  ? 

2.  Why  do  we  not  take  the  origin  for  the  center  of  the  system  of  circles 
in  proving  Theorem  X  ? 

3.  Construct  a  number  of  lines  of  the  system  x  =  a  and  the  inverse  circles. 


308 


ANALYTIC  GEOMETRY 


4.  Construct  a  number  of  lines  of  the  system  y  =  mx  +  I  and  the 
inverse  circles. 

5.  Construct  a  number  of  circles  of  the  system  x^  -{-  y^  —  Sx  i- 1  —  r^  =  0 
and  the  inverse  circles. 

6.  What  is  the  inverse  of  a  system  of  tangent  circles  if  the  point  of  tan- 
gency  is  the  center  of  inversion  ? 

7.  What  is  the  inverse  of  a  system  of  circles  passing  through  two  points 
one  of  which  is  the  center  of  inversion  ? 

8.  What  is  the  inverse  of  a  system  of  circles  with  two  limiting  points 
one  of  which  is  the  center  of  inversion  ? 

9.  The  point  Pi(xi,  yi)  may  be  regarded  as  a  point-circle  whose  equa- 
tion is  (x  —  Xi)2  +  (y  —  ?/i)2  =:  0.  Show  that  the  system  of  circles  repre- 
sented by  (x  —  Xi)2  +  (y  —  y\Y  -f  A;  (x^  +  y^  —  1)  =  0  has  two  limiting  points, 
namely,  Pi  and  the  inverse  of  Pi.  What  is  the  nature  of  the  system  if  Pi 
lies  on  the  circle  x^  +  y^  =  1  ? 

10,  How  may  problem  9  be  used  to  define  an  inversion  ? 

130.  Orthogonal  systems  of  circles.  Two  systems  of  circles  are  said  to  be 
orthogonal  if  each  circle  of  one  system  is  orthogonal  (p.  143)  to  every  circle 
of  the  other  system.  The  preceding  sections  enable  us  to  construct  such 
systems  with  ease. 


Consider  two  systems  of  parallel  lines  such  that  the  lines  of  one  system 
are  perpendicular  to  the  lines  of  the  other.  If  we  invert  these  systems  of 
lines,  we  get  two  systems  of  tangent  circles  whose  centers  lie  respectively  on 
two  perpendicular  lines  (Theorem  VIII,  p.  306).  Since  angles  are  preserved 
by  inversion  (Corollary,  p.  306)  these  systems  are  orthogonal.     Hence 


i 


INVERSION 


809 


Theorem  XI.  Two  systems  of  tangent  circles  are  orthogonal  if  they  have  the 
same  point  of  tangency  and  if  their  centers  lie  on  perpendicular  lines. 

It  is  also  evident  that  all  of  the  lines  passing  through  the  same  point  P 
and  all  of  the  circles  having  the  center  P  intersect  orthogonally.  The 
inverse  of  the  system  of  lines  is  the  system  of  circles  passing  through 


the  origin,  and  the  inverse  of  P  (Theorem  IX^  p.  306),  and  the  inverse  of 
the  system  of  concentric  circles  is  the  system  of  circles  having  the  origin 
and  the  inverse  of  P  as  limiting  points  (Theorem  X,  p.  307).     Hence 

Theorem  XII.     Two  systems  of  circles  are  orthogonal  if  all  the  circles  of  one 
system  pass  through  two  points  which  are  the  limiting  points  of  the  other. 


MISCELLANEOUS  PROBLEMS 

1.  Show  that  the  degree  of  an  equation  is,  in  general,  doubled  by  an 
inversion.  Will  this  be  true  if  the  terms  of  the  highest  ^degree  contain 
x2  +  2/2  as  a  factor  ? 

2.  Construct  a  linkage  consisting  of  a  deformable  rhombus  APBP'  and 
two  bars  of  equal  length  OA  and  OB  which  are  free  to  rotate  about  the  fixed 
point  0.  Show  that  P  and  P'  describe  inverse  curves  if  0  is  the  center  of 
inversion  and  OA^  —  AP^  is  the  unit  of  length. 

3.  If  P  is  that  point  of  the  rhombus  in  problem  2  which  lies  nearest  to  0, 
then  by  adding  a  bar  O'P,  which  is  free  to  rotate  about  the  fixed  point  0', 
P  will  be  constrained  to  move  in  a  circle.  How  will  P'  move  ?  This  linkage 
is  known  as  Peaucellier's  Inversor. 

4.  Show  how  to  construct  four  circles  passing  through  a  given  point  and 
tangent  to  each  of  two  given  circles  which  do  not  intersect. 

Hint.    Invert  the  figure,  using  the  given  point  as  the  center  of  inversion. 

5.  Find  the  properties  of  the  cissoid,  lemniscate,  strophoid,  cardioid,  and 
limaQon,  which  may  be  obtained  from  problems  3,  4,  5,  6,  9,  10,  12,  and  13, 
p.  220,  by  inversion  with  a  proper  center. 

6.  Show  that  the  angle  which  one  line  makes  with  a  second  equals  the  angle 
between  the  inverse  circles,  without  using  the  Corollary  on  p.  305. 


CHAPTER  XY 


POLES  AND  POLARS.    POLAR  RECIPROCATION 


131.  Pole  and  polar  with  respect  to  a  circle.   Let  Pi(xi,  yi)  be  any  point 
and  let  the  equation  of  a  given  circle  C  be 

(1)  a;2  +  2/2  =  r2. 
The  line  ii,  whose  equation  is 

(2)  a?iX  +  i/i2/  =  ^^ 

is  called  the  polar  of  Pi  (xi,  yi)  with  respect  to  C,  and  Pi  is  called  the  pole  of  Xi. 
Theorem  I.    The  polar  of  a  point  on  a  circle  is  the  tangent  to  the  circle  at 
that  point 

The  proof  follows  at  once  from  the  definition  and  from  the  fact  that  (2)  has  the  same 
form  as  the  equation  of  the  tangent  (Theorem  T,  p.  212). 

Theorem  II.    The  polar  of  a  point  Pi  vjith  respect  to  a  circle  is  perpendicular 
to  the  line  passing  through  Pi  and  the  center  of  the  circle. 

Proof.    Thp  equation  of  the  line  passing  through  Pi  and  the  origin,  the 
center  of  the  circle  (1),  is  (Theorem  VII,  p.  97) 

yix  -  Xiy  =  0. 

This  line  is  perpendicular  to  (2),  the  polar  of  Pi  (Corollary  III,  p.  87). 

Q.E.D. 

Corollary.    The  angle  formed  by  the  polar s  of  two  points  with  respect  to  a 
circle  is  equal  to  the  angle  formed  by  the  lines  joining  those  points  to  the  center 

of  the  circle. 

Theorem  III.    The  polar  of  any  point  of  a 

given  line  passes  through  the  pole  of  that  line. 

Proof    Let  Li  be  the  given  line  and  let 

Pi  (-^1,  yi)  l)e  its  pole.     Then  the  equation 

of  ii  is 

(3)  xix  +  yiy  =  r^. 

Let  P2  (X2, 2/2)  he  any  point  on  ii;  then 
(Corollary,  p.  53) 

(4)  X1X2  +  yiyz  =  r2. 
The  equation  of  the  polar  L2  of  the  point 

P2is 

X2X  +  7/oy  =  r2. 
310 


POLES  AND  POLARS 


311 


This  line  passes  through  Pi,  for  if  the  coordinates  of  Pi  be  substituted 
for  X  and  y,  we  obtain  equation  (4),  which  is  known  to  be  true. 

Corollary.  The  pole  of  any  line  is  the  point  of  intersection  of  the  polars  of 
any  two  of  its  points. 

Theorem  IV.  The  pole  of  any  line  passing  through  a  given  point  lies  on  the 
polar  of  that  point. 

Proof.    Let  Pi  (xi,  yi)  be  the  given  point.     Its  polar  is  the  line 

(5)  Xi :  Xix  +  yiy  =  r^. 

Let  P2(iC2,  2^2)  l>e  the  pole  of  a  line  L2  which  passes  through  Pi.'  The 
equation  of  L^  is  then 

X2X  +  2/22/  =  r^- 

Since  L2  passes  through  Pi  we  have  (Corollary,  p.  53) 

(6)  xgxi  +  y^yi  =  r2. 

Then  P2  lies  on  ii,  for  when  the  coordinates  of  P2  are  substituted  in  (5) 
for  X  and  y,  we  obtain  equation  (6),  which  is  known  to  be  true.  q.e.d. 

Corollary.  The  polar  of  any  point  is  the  line  passing  through  the  poles  of 
any  two  lines  which  pass  through  the  given  point. 

132.  Construction  of  poles  and  polars. 

Construction  I.  To  construct  the  polar  of  a  point  P  outside  of  a  circle, 
draw  the  tangents  to  the  circle  which  pass  through  P.  The  line  joining  the 
points  of  contact  of  these  tangents  is  the  polar  of  P. 

Proof.  Let  Xi  and  Lo  be  the  tan- 
gents to  0,  and  let  Pi  and  P2  be  their 
points  of  tangency.  Then  the  polars 
of  Pi  and  P2  are  the  lines  ii  and  L2 
(Theorem  I).  Since  ii  and  L2  pass 
through  P,  the  polar  of  P  is  the  line  L 
passing  through  Pi  and  P2  (Corollary 
to  Theorem  IV).  q.e.d. 

In  like  manner  the  following  con- 
structions are  proved. 

Construction  II.  To  construct  the  pole 
>f  a  line  L  which  cuts  the  circle,  draw 
le  tangents  at  the  points  at  which  L 
itersects  the  circle.     The  point  of  inter- 
action of  these  tangents  is  the  pole  of  L  (Corollary  to  Theorem  III). 


312 


ANALYTIC  GEOMETKY 


Construction  III.    To  construct  the  polar  of  a  point  P  within  a  circle, 

construct  the  poles  Pi  and  P2  of 
two  lines  Xi  and  L^  passing  through 
P  (Construction  II).  The  line  join- 
ing Pi  and  P2  is  the  polar  of  P 
(Corollary  to  Theorem  IV). 


Construction  IV.  To  construct 
the  pole  of  a  line  L  which  does  not 
cut  the  circle,  construct  the  polars 
Xi  and  L2  of  two  points  Pi  and  P^ 
on  L  (Construction  I) .  The  inter- 
section of  ii  and  L2  is  the  pole  of 
L  (Corollary  to  Theorem  III). 


Y' 

- 

\ 

X'    {          0 

^          X 

1  ^ 

rC^ 

\ 

^ 

\  I 

r' 

-T-- 

PROBLEMS 

19 

1.  Find  the  equation  of  the  polar  of  each  of  the  following  points  with 
respect  to  the  given  circle  and  construct  the  figure. 

(a)  (3,  _  4),  x2  +  2/2  =  4.  (d)  (3,  4),       x^  +  ?/2  =  36. 

(b)  (-  1,  2),  x2  +  y2  ^  25.  (e)  (5,  0),       x^  +  y^  =  49. 

(C)    (7,    -  2),    X2  +  2/2  ^  9.  (f)    (_  3,  4),  x2  +  2/2  =  25. 

2.  Find  the  pole  of  each  of  the  following  lines  with  respect  to  the  given 
circle  and  construct  the  figure. 

(a)  3  x  +  ?/  =  25,  £c2  +  y2  ^  25.  Ans.  (3,  1). 

(b)  3x  -  2?/  =  18,  x2  +  2/2  ^  36.  Ans.  (6,  -  4) 

(c)  x-4y +  8  =  0,  x2  +  ?/2  =  16.  Ans.  (-2,8) 

(d)  2x-y  =  64,  x2  +  2/^  =  64.  Ans.  (2,  -  1) 

(e)  x-32/  +  16  =  0,  x2  +  2/2  =  16.  Ans.  (-1,3) 
if)  x-^y  =  18,  x2  +  2/2  ^  9.  Ans.  (i,  -  |) 

(g)  Ax  +  By  +  C  =  ^,  x2  +  2/2  =  r^.  Ans.  (^-^ 


J5r2\ 


Hint.  Let  Pi(«i,  ?/i)  be  the  pole  of  the  given  line  and  write  down  the  equation  of  the 
polar  of  Pt  with  respect  to  the  given  circle.  From  the  conditions  that  this  line  shall 
coincide  with  the  given  line  (Theorem  III,  p.  88)  determine  x^  and  y^. 

3.  Find  the  distance  from  the  origin  to  the  polar  of  Pi  with  respect  to 


x2  +  2/2  =  r2. 


Ans. 


Vxi2  + 


yi' 


4.  By  problem  3  show  that  (a)  if  Pi  approaches  the  origin  its  polar  recedes 
to  infinity;  (b)  if  Pi  recedes  to  infinity  its  polar  approaches  the  origin. 


POLES  AND  POLARS 


813 


6.  By  problem  2,  (g),  show  that  if  a  line  recedes  to  infinity  its  pole 
approaches  the  origin,  and  if  the  line  approaches  the  origin  its  pole  recedes 
to  infinity. 

6.  Find  the  pole  of  the  line  joining  Pi  {xi,  y\)  and  P2  (X2,  2/2)  and  prove 
that  it  is  the  point  of  intersection  of  the  polars  of  Pi  and  P2. 

\X1y2  -  iczz/i'  Xiy2  -  x^yj 

7.  Find  the  polar  of -the  point  of  intersection  of  AiX  -f  Biy  +  Ci  =  0  and 
A<iX  +  B^y  +  C2  =  0  and  prove  that  it  passes  through  the  poles  of  these  lines. 

Ans.    {Bid  -  B2C1)  X  +  {C1A2  -C2A1)  y  =  {A1B2  -  A2B1)  r^. 

8.  If  the  line  y  —  yi  =  m{x  —  Xi)  revolves  about  Pi,  the  locus  of  its  pole 
is  the  polar  of  Pi. 

9.  The  radius  of  a  circle  is  a  mean  proportional  between  the  distances 
from  its  center  to  any  point  and  to  the  polar  of  that  point. 


133.  Polar  reciprocation  with  respect  to  a  circle.  The  transformation 
which  replaces  a  given  line  by  its  pole  with  respect  to  a  circle  is  called  a 
polar  reciprocation  with  respect  to  that  circle.  Analytically  the  transforma- 
tion is  defined  by 

Theorem  V.    The  pole  of  the  line 

Ax  +  By  +  C  =  0 
with  respect  to  the  circle  x^  4-  ?/2  =  r^ 


is  the  point 


Proof.    Let  Pi  (xi,  yi)  be  the  pole  of  the  given  line.     Then  the  polar  of 
Pi  is  the  line 

XiX  +  yiy  -  r2  =  0. 

Then,  by  Theorem  III,  p.  88,  X 


Xi  = 


xi^yi 
A  B 
Ar^ 
C  ' 


_  _r2 
C 

2/1  =  - 


Q.E.D. 


The  locus  C  of  the  poles  of  the  tangents 
to  a  curve  C  is  called  the  polar  reciprocal 
of  C. 


314 


ANALYTIC  GEOMETRY 


Theorem  VI.    If  C  is  the  polar  reciprocal  of  a  curve  C,  then  C  is  the  polar 
reciprocal  of  C\ 

Proof.    Let  I  and  m  be  two  tangents  to  C  at  i  and  M.     Let  M'  be  the 

pole  of  m  and  L'  of  I.  Then  L'  and  M'  are 
two  points  on  C  by  definition. 

Let  p^  be  the  line  passing  through  L' 
and  M\  Then  the  pole  of  p'  is  P,  the 
point  of  intersection  of  I  and  m  (Corollary 
to  Theorem  III,  p.  311). 

Let  L'  move  along  C"  until  it  comes  into 
coincidence  with  if.  Then  the  limiting 
position  of  p'  is  the  tangent  to  C  at  M'. 
But  as  i' approaches  M%  I  must  approach  m, 
and  the  limiting  position  of  P  is  evidently 
the  point  M.  Hence  M  is  the  pole  of  the 
tangent  to  C  at  Jf'.     Hence  O  is  the  polar  reciprocal  of  C".  q.e.d. 


The  method  of  finding  the  equation  of  C  from  that  of  C  is  illustrated  by 

Ex.  1.  Find  the  polar  reciprocal  of  the  parabola  2/^=  4cc  with  respect  to  the 
circle  x^  -\-y'^  =  4. 

Solution.   Let  Pi  (a^i,  yi)  be  any  point  on  the  parabola.    Then 

(1)  2/i2=4ki. 

The  equation  of  the  tangent  to  the  parabola 
at  Pi  is  (Theorem  III,  p.  214) 


(2) 


yiy^2{x  +  x{),  or 
2  X  -  2/11/  +  2  a^i  ==  0. 


By  Theorem  V,  the  pole  of  (2)  is  the  point 
P'  {x'f  2/0  >  where 


x'  = 

4 

Xi 

y'  = 

2  2/1 

Xi 

Hence 

Xi=- 

4 

x'' 

yi  = 

2  2/' 

X' 

Substituting  in 

(1), 

y'-2  = 

-4a' 

> 

Vs 

— 

-1^ 

— 

h 

^ 

^ 

V 

s. 

^1 

>^ 

x 

' 

^ 

^ 

•' 

sc 

7 

^ 

>* 

\ 

X 

^ 

X 

0 

^ 

X 

^s 

/ 

S. 

y 

k 

^ 

y' 

S 

X 

^ 

1 

-1- 

^ 

1 

This  is  the  equation  of  the  locus  of  P',  that  is,  of  the  polar  reciprocal  of  the 
given  parabola.  The  polar  reciprocal  is  therefore  a  parabola  of  the  same  size, 
turned  to  the  left  instead  of  to  the  right. 

The  method  consists  in  finding  the  pole  P'  of  the  tangent  to  the  given  curve  at 
Pi,  expressing  xi  and  2/1  in  terms  of  x'  and  y^,  and  substituting  in  the  given 
equation. 


POLES  AND  POLARS  315 

PROBLEMS 

1.  Find  the  polar  reciprocal  of  each  of  the  following  circles  with  respect 
to  the  circle  x^  -^  y^  =  4. 

(a)  x2  +  2/2  _  4 x  =  0.  Ans.   y^  +  4x  =  ^. 

(b)  x2  +  2/2  -  2  X  -  3  =  0.  Ans.    3  x2  +  4  y2  +  g  x  -  16  =  0. 

(c)  x2  +  2/2  -  6  X  +  5  =  0.  Ans.    5  x2  -  4  2/2  -  24  x  +  16  =  0. 

2.  Find  the  polar  reciprocal  of  each  of  the  following  curves  with  respect 
to  the  given  circle. 

(a)  x2  +  4  2/2  =  16,  x2  +  2/2  =  1.  Ans.    16  x2  +  4  2/2  =  1. 

(b)  2/^  =  2x -6,  x2  +  2/^  =  9.  Ans.    6x2  -  2/2  -  18x  =  0. 

(c)  4x2  +  2/2  =  8x,  x2  +  2/2  =  4.  Ans.   2/^  +  2x  -  4  =  0. 

3.  Verify  the  answers  to  problems  1  and  2  by  finding  the  polar  recipro- 
cals of  the  curves  given  in  the  answers  and  applying  Theorem  VI. 

4.  Show  that  the  equilateral  hyperbola  2  X2/  =  9  is  transformed  into  itself 
by  a  polar  reciprocation  with  respect  to  the  circle  x^  +  y^  =  9. 

5.  Show  that  the  locus  of'x2  —  y"^  =  a'^  is  transformed  into  itself  by  a  polar 
reciprocation  with  respect  to  the  circle  x^  -{-  y^  =  a^. 

134.  Pole  and  polar  with  respect  to  the  locus  of  any  equation  of  the 
second  degree.  Let  Pi  (xi,  2/1)  be  any  point  and  let  any  equation  of  the 
second  degree  be 

(1)  ^x2  +  Bxy  ■i-Cy^-h  Dx  +  Ey  +  F=0. 

The  line  Xi,  whose  equation  has  the  same  form  as  the  tangent,  namely 
(Theorem  II,  p.  212), 

(2)  A...+B^i^^^  +  Cp.y+1>'^  +  B^  +  F=0, 

is  called  the  polar  of  the  point  Pi  with  respect  to  the  locus  of  (1)^     Pi  is 
called  the  pole  of  Li. 

In  what  follows  we  speak  always  of  poles  and  polars  with  respect  to 
the  locus  of  (1)  unless  the  contrary  is  stated.  The  following  theorems  are 
generalizations  of  the  theorems  indicated,  and  are  proved  in  the  same  way 
by  using  (1)  and  (2)  of  this  section  instead  of  (1)  and  (2)  of  section  131, 
p.  310. 

Theorem  VII.  (Generalization  of  Theorem  I.)  The  polar  of  a  point  on  the 
locus  of  (1)  is  the  tangent  at  that  point. 

Theorem  VIII.  (Generalization  of  Theorems  III  and  IV.)  The  polar  of  any 
point  on  a  given  line  passes  through  the  pole  of  that  line.  Conversely.,  the  pole 
of  any  line  passing  through  a  given  point  lies  on  the  polar  of  that  point. 


316  ANALYTIC  GEOMETRY 

Corollary  I.  The  pole  of  any  line  is  the  point  of  intersection  of  the  polars  of 
any  two  of  its  points. 

Corollary  II.  The  polar  of  any  point  is  the  line  passing  through  the  poles  of 
any  two  lines  which  pass  through  the  given  point. 

The  constructions  on  pp.  311  and  312  enable  us  to  construct  poles  and 
polars  with  respect  to  (1),  for  the  theorems  by  which  the  constructions  are 
proved  have  been  generalized  for  the  locus  of  (1). 

A  good  idea  of  the  direction  of  the  polar  of  a  point  with  respect  to  a  conic  is 
afforded  by 

Theorem  IX.  The  polar  of  a  point  Pi  with  respect  to  a  conic  is  parallel  to  the 
tangent  to  the  conic  at  the  point  where  the  diameter  through  Pi  cuts  the  conic. 

Proof.  The  proof  is  separated  into  two  cases  according  as  the  conic  is  a  central 
conic  or  a  parabola. 

Case  I.   Central  conic.   If  the  center  is  the  origin,  its  equation  may  be  written 

Ax'^ -{■  Cy2 -{- F  =  0. 
The  equation  of  the  polar  of  Pi  is  \P 

(3)  Axix  +  Cyiy  +  F  =  0. 

Let  the  diameter  through  Pi  cut  the  conic  at  P2.  The  equation  of  the  tangent 
at  P2  is 

(4)  Ax2X  +  Cyip  +  F  =  0. 

Since  Pi  and  P2  are  on  a  line  through  the  origin  (Corollary,  p.  242), 

xi_yi^ 

and  hence  the  lines  (3)  and  (4)  are  parallel  (Corollary  II,  p.  87). 

,  Case  II.    Parabola.    Its  equation  is  y^  =  2px. 
The  equation  of  the  polar  of  Pi  is 

(5)  yiy^p{x-\-^\)- 

Let  the  diameter  through  Pi  cut  the  parabola  at  P2.  The  equation  of  the 
tangent  at  P2  is 

(6)  2/22/=p(«  +  iC2). 

Since  (Theorem  X,  p.  242)  y^  =  ?/2,  the  lines  (5)  and  (6)  are  parallel.         q.e.d. 

PROBLEMS 

1 .  Find  the  equations  of  the  polars  of  the  following  points  with  respect  to 
the  given  conies  and  construct  the  figures. 

(a)  (3,  4),      9x2  +  4y2  =  36.         (e)  (- 1,  3),  a;2  +  a;?/ -  6y  +  4  =  0. 

(b)  (2,  -  1),  16  x2  -  ?/2  =  64.  (f )  (4,  5),       xy  -{- ^x  -  Qy  -^  =  0. 

(c)  (3,  6),       x2  +  4  y  =  0.  (g)  (2,-6),    x'^  +  'lxy  ^y^ +  x -y  =  0. 

(d)  (2,  -  4),  xy  -\Q  =  0.  (h)  (3,  2),       5x2  +  6a:?/  +  5?/2 - 12  =  0. 


POLES  AND  POLARS  317 

2.  Find  the  poles  of  the  following  lines  with  respect  to  the  given  conies 
and  construct  the  figures. 

(a)  9x  +  4y  =  36,  9x2  +  ^/^  =  36.  Ans.  (1,  4). 

(b)  2x-3y +  4  =  0,  2/2  =  4x.  Ans.  {2,  -  S). 

(c)  x-2y  =  16,  xy  =  8.  Ans.  (-  2,  1). 

(d)  14x  +  2/  =  8,  4x2  -  ?/2  =  16.  Ans.  (7,  -  2). 

(e)  2x  -  y  +  13  =  0,  x2  4-  4?/  =  16.  Ans.  (-  4,  -  6). 

(f )  X  +  4  =  0,  x2  +  4  xy  +  y2  +  2  X  +  4  =  0.  Ans.  (0,  0). 

(g)  llx  +  2y +  18  =  0,  17x2-12x?/ +  82/2-68x  +  24y  -  12  =  0. 

Ans.  (0,  -2). 

3.  Tangents  are  drawn  from  the  point  (8,  4)  to  the  ellipse  x2  +  4^/2  =  16. 
Find  the  equation  of  the  line  joining  their  points  of  tangency. 

Ans.    x  +  2?/-2  =  0. 

4.  Tangents  are  drawn  to  the  hyperbola  16x2  _  y2  —  64  at  the  points 
of  intersection  of  the  hyperbola  and  the  line  8x  +  3?/  +  32  =  0.  Find  the 
coordinates  of  their  point  of  intersection.  Ans.  (—1,^). 

5.  How  does  the  polar  of  a  point  with  respect  to  a  central  conic  behave 
if  the  point  approaches  the  center  ?  if  the  point  recedes  to  infinity  ? 

6.  The  polar  of  the  focus  of  any  conic  with  respect  to  that  conic  is  the 
corresponding  directrix. 

7.  The  polar  of  any  point  on  the  directrix  of  a  conic  passes  through  the 
corresponding  focus. 

8.  The  polars  of  a  point  with  respect  to  conjugate  hyperbolas  are  parallel. 

9.  The  polar  of  a  focus  of  an  ellipse  with  respect  to  the  major  auxiliary 
circle  is  the  corresponding  directrix. 

10.  What  is  the  locus  of  a  point  which  lies  on  its  polar  with  respect  to  a 
given  conic  ? 

1 1 .  That  part  of  the  diameter  of  a  parabola  included  between  any  point 
on  it  and  its  polar  is  bisected  by  the  point  of  contact. 

135.  Polar  reciprocation  with  respect  to  the  locus  of  any  equation  of  the 
second  degree.  Let 

(1)  ^x2  +  Bxy  +  Cy^  +  Dx  +  Ey  +  F=0 

be  any  equation  of  the  second  degree.  Let  C  be  any  curve  and  let  C  be  the 
locus  of  the  poles  of  the  tangents  to  C  with  respect  to  the  locus  of  (1).  C  is 
called  the  polar  reciprocal  of  C  with  respect  to  (1). 

Theorem  X.  (Generalization  of  Theorem  VI.)  If  C  is  the  polar  reciprocal 
of  C  with  respect  to  (1),  then  C  is  the  polar  reciprocal  of  C. 

The  proof  is  identical  with  that  of  Theorem  VI,  p.  314.  For  the  theorems  on  poles 
and  polars  with  respect  to  a  circle,  used  in  proving  that  theorem,  have  been  extended  to 
the  locus  of  (1). 


318 


ANALYTIC  GEOMETRY 


Corollary.  The  polar  reciprocal  of  C  is  a  curve  C  whose  tangents  are  the 
polar s  of  the  points  of  C. 

The  polar  reciprocal  of  a  curve  C  with  respect  to  (1)  may  therefore  be 
regarded  in  either  one  of  two  ways : 

1.  As  the  locus  of  the  poles  of  the  tangents  to  C. 

2.  As  the  curve  whose  tangents  are  the  polars  of  the  points  of  C. 

In  either  case  the  fact  to  be  observed  is  that  to  a  point  of  one  figure  corre- 
sponds a  straight  line  of  the  other  figure  and  vice  versa.  The  transformation 
which  replaces  C  by  C  is  called  a  polar  reciprocation  with  respect  to  (1). 

Analytically  the  polar  reciprocation  with  respect  to  (1)  is  completely  defined 
by  the  equation 


2  -"2  2 

For,  in  the  first  place,  the  locus  of  (2)  gives  us  at  once  the  polar  of 

Pi  i^u  yi)' 

In  the  second  place,  the  pole  of  any  line 
(3)  A'x  -\-  B'y  -{■  C  =  0  > -> 

is  found  from  (2)  as  follows.  Let  Pi  (xi,  2/1)  be  the  pole  of  (3).  Then  since 
(2)  and  (3)  are  the  equations  of  the  polar  of  the  same  point  Pi,  their  loci 
coincide.     Hence  (Theorem  III,  p.  88) 


^Xi  +  -^1  +  - 


-x,+  Cy,  +  - 


D  E  ^ 

-Xi  +  -yi  +  P 


A'  B'  C 

These  equations  can,  in  general,  be  solved  for  Xi  and  yi  (Theorem  IV,  p.  90). 
The  .method  of  finding  the  equation  of  the  polar  reciprocal  of  a  given 
curve  C  is  illustrated  in  the  following  example. 

Ex.  1.  Find  the  equation  of  the  polar  reciprocal  of  the  ellipse 

(7:4x2  +  92/2-1  =  0 
with  respect  to  the  ellipse 

(4)  a;2  +  4?/2  +  2x  =  0. 
Solution.   Let  Pi  (xi,  yi) 

be  any  point  on  C.    Then 

(5)  4a;i2  +  9?/i2-l  =  0. 
The  equation  of  the  tan- 
gent to  C  at  Pi  is  (Theorem 
III,  p.  214) 

(6)  4:xix +  9 yiy -1  =  0. 
Let  P'{x\  y')  be  the  pole 

of  (6)  with  respect  to  (4) .  The 
equation  of  the  polar  of  P'  is 

(7)  {x'-\-l)x  +  ^y'y  +  x'=0. 


s». 

y 

^ 

ZP 

x. 

y 

^. 

y 

r      25" 

\    t                 > 

^  V       3^-. 

r  t           i 

--— z^ 

^ 
V 

Z 

S 

.^ 

"S. 

^'^ 

^s,^ 

-r-  — 

^;: 

POLES  AND  POLARS  319 


Since  (6)  and  (7)  have  the  same  locus  (Theorem  III,  p.  88), ' 

4a;i    ^^yi^  -1, 
x'  +  l      4  y'        x' 
Solving  for  xi  and  yi,  we  obtain 

Substituting  in  (5) ,  we  have  the  required  equation 

Reducing  and  dropping  primes,  we  obtain 

27x2  _64y2_i8x- 9  =  0, 
whose  locus  is  an  hyperbola. 

In  the  figure  three  divisions  are  taken  for  unity. 

PROBLEMS 

1.  Find  the  polar  reciprocal  of  the  first  of  the  following  curves  with 
respect  to  the  second.     Construct  the  figure  in  each  case. 

(a)  y2  _  4  a;  =  0,  x^  +  iy  =  0.  Ans.  xy  -2  =  0. 

(b)  ic2  +  ?/2  =  1,  x^-y^  =  4.  Ans.  x^ -{- y^  =  16. 

(c)  x2  +  4?/2  =  4,  4x2  +  y2  =  4.  j^y^s.  64x2  +  2/2  =  16. 

(d)  x2 -42/2  =  16,  x2  + 4^2:^  2x.  Ans.  16x2  -  64  2/2  -  32x  +  16  =  0. 

(e)  xy  -  4  =  0,  x^  -  y^  =  16.  Ans.  xy  +  16  =  0. 

(f)  8  2/  -  x3  =  0,  x2  -  2/2  =  4.  Ans.  2  x^  =  27  y. 

2.  Verify  the  answers  to  problem  1  by  showing  that  the  polar  reciprocals 
of  the  curves  in  the  answers  are  the  given  curves. 

3.  Show  that  either  of  the  following  curves  is  unchanged  by  a  polar  recip- 
rocation with  respect  to  the  other. 

(a)  62x2  +  a22/2  =  aW,  hH'^  -  a^y'^  =  a^"^. 

(b)  &2x2  _  a^y2  -  c^252^  12x2  _  a^y2  ^  _  ^252. 

(c)  2/2  -  2px  =  0,  2/2  +  2px  =  0. 

4.  If  the  vertices  of  one  triangle  are  the  poles  of  the  sides  of  a  second 
triangle,  then  the  vertices  of  the  second  are  the  poles  of  the  sides  of  the  first. 

Two  triangles  such  that  the  vertices  of  either  are  the  poles  of  the  sides  of 
the  other  are  called  conjugate  triangles.  If  the  vertices  of  a  triangle  are  the 
poles  of  the  opposite  sides,  the  triangle  is  said  to  be  self-conjugate. 

5.  Show  that  (2,  1),  (4,  4),  and  (3,  2)  are  the  vertices  of  a  self-conjugate 
triangle  with  respect  to  the  hyperbola  x^  —  y^  =  4. 

6.  Show  how  to  construct  a  self -con  jugate  triangle  with  respect  to  a  given 
conic  if  one  vertex  is  given.     How  many  may  be  constructed  ? 


320 


ANALYTIC  GEOMETRY 


7.  Show  that  if  we  reciprocate  the  figure  which  is  given  or  implied  in  one 
of  the  following  statements,  we  obtain  the  corresponding  statement. 


(a)  Two  points  determine  the  line 
on  which  they  lie. 

(b)  Three  points  on  the  same  line. 

(c)  Three  points  at  the  vertices  of  a 
triangle. 

(d)  n  points  at  the  vertices  of  a  poly- 
gon. 

(e)  An  infinite  number  of  points 
lying  on  a  curve. 

(f)  A  line  intersecting  a  curve  in  n 
points. 

(g)  A  curve  passing  twice  through 
the  same  point. 

(h)  A  conic  section. 

(i)  A  conic  may  be  constructed  which 
passes  through  five  given  points. 

(j)  Two  conies  intersect  in  general 
in  four  points. 


Two  lines  determine  the  point  in 
which  they  intersect. 

Three  lines  through  the  same  point. 
Three  lines  forming  a  triangle. 

n  lines  forming  the  sides  of  a  poly- 


gon. 

An  infinite  number  of  lines  tangent 
to  a  curve. 

A  point  through  which  pass  n  lines 
tangent  to  a  curve. 

A  curve  tangent  twice  to  the  same 
line. 

A  conic  section. 

A  conic  may  be  constructed  which 
is  tangent  to  five  given  lines. 

Two  conies  have  in  general  four 
common  tangents. 

136.  Polar  reciprocation  of  a  circle  with  respect  to  a  circle.    The  equa- 
tions of  any  two  circles  C  and  Oi  may  be  put  in  the  forms 

C  :  a:2  +  y2  ^  r2 
and  Ci :  x2  +  ?/2  +  Zte  +  i?'  =  0 

by  taking  the  center  of  C  as  origin  and  the  line  of  centers  of  C  and  C\  as 
the  X-axis,     We  shall  now  find  the  polar  reciprocal  of  C  with  respect  to  C\. 

Let  Pi  (xi,  2/i)  be  any  point  on  C.     Then  (Corollary,  p.  53) 

(1)  Xi2  +  yi2  =  r\ 

and  the  equation  of  the  tangent  to  C  at  Pi  is  (Theorem  I,  p.  212) 

(2)  xix  +  yiy  -  r2  =  0. 

Let  P'  (x',  y")  be  the  pole  of  (2)  with  respect  to  Ci ;  then  the  polar  of  P'  is 


X'X  ^y'y^B  ?^^  +  F 


OX 

(3) 


x'  + 


-)x  +  2/'2/  +  -x'  +  P 


0, 


0. 


Since  (2)  and  (3)  have  the  same  locus  (Theorem  III,  p. 

xx    _yi_    -y^    ^ 
y' 


Solving  for  Xi  and  yi,  we  obtain 
r2  (2  x'  +  D) 


x'  +  F 


Xi  = 


Bx'  +  2F 


y\  = 


2r^y' 


Dx'  -\-2F 


POLES  AND  POLARS 


321 


Substituting  these  values  in  (1),  reducing,  and  dropping  primes,  we  have 
the  equation  of  C\  namely, 

C  :  (4r2  -  2)2) x2  +  4r22/2  +  4i)(r2  -F)x-\-  {r^B^  -  42^2)  _  o. 

The  discriminant  of  C  is  (p.  265) 
0'  =  16  r2  (4  r2  -  2)2)  (rW^  -iF^)-  64  rW^  (r^  -F)^  =  -167^  (2)2  -  4  2?')2. 


As  ^  v2)2  -  4  2^  is  the  radius  of  Ci  (Theorem  I,  p.  131),  it  follows  that 
0'  is  not  zero  if  the  radii  of  C  and  Ci  are  not  zero.  Hence  (Theorem  I, 
p.  266) 

Theorem  XI.    The  polar  reciprocal  of  the  circle  C  :  x^  +  y^  =  r^  with  respect 
to  the  circle  Ci :  x^  +  y^  +  Dx  -\-  F  =  0  is  the  non-degenerate  conic  C  whose 
equation  is 
(XI)  (4r2  -  2)2)a;2  +  4r22/2  +  42)(r2  -  F)x  +  (r22)2  -  42^2)  =  q. 

The  nature  of  the  conic  C  depends  upon  the  sign  of 
A'  =  _  4  . 4  r2  (4  r2  -  2)2). 

It  is  evident  that 

2)2 
A'  <  0  if  4  r2  -  2)2  >  0,  or  r2  >  — ; 

4 

2)2 

A'  >  0  if  4  r2  -  2)2  <  0,  or  r2  <  — ; 

4 

A'  =  0  if  4  r2  -  2)2  =  0,  or  r2  =  — . 


Hence  (Theorem  IX,  p.  277) 

2)2 

the  conic  C  is  an  ellipse       if  r2  >  — ; 

4 


the  conic  C  is  an  hyperbola  \ir^  <—^ 
the  conic  C"  is  a  parabola     if  r2 


4 
4 


2)2  /      2)       \ 

But  —  is  the  square  of  the  distance  from  the  origin  to  (  —  — ,  0  ) ,  the 
4  \      2       / 


center  of  Ci  (Theorem  I,  p.  131),  and  therefore 

2)2 
the  center  of  Ci  is  inside  of  C   if  r2  >  — ; 

4 

2)2 

the  center  of  Ci  is  outside  of  C  if  r2  <  —  ; 

4 


and 


the  center  of  Ci  is  on  C 


2)2 

ifr2=— . 
4 


322  ANALYTIC  GEOMETRY 

Hence 

Theorem  XII.  The  polar  reciprocal  of  a  circle  C  with  respect  to  a  circle  C\ 
is  an  ellipse,  hyperbola,  or  parabola  according  as  the  center  of  Ci  is  inside  of, 
outside  of,  or  on  the  circle  C. 

PROBLEMS 

1.  Find  the  polar  reciprocal  of  the  circle  x^  -\-  y^  =  i  with  respect  to  each 
of  the  following  circles  and  construct  the  figure. 


(a)  x2  +  2/2-4x-5  =  0. 

Ans. 

4?/2-36iK-9  =  0. 

(b)  x2  +  2/2_2a;-;3:=0. 

Ans. 

3x2 +  42/2 -Ux- 5  =  0. 

(c)  x'^  +  y^-6x  =  0. 

Ans. 

5x2  _  42/2 +  24x- 36  =  0. 

2.  Show  that  the  center  of  Ci  (Theorem  XI)  is  a  focus  of  (XI)  and  that  the 
corresponding  directrix  is  the  polar  of  the  center  of  C  with  respect  to  Ci. 

Hint.  Transform  (XI)  by  moving  the  origin  to  the  center  of  C^,  find  the  focus  and 
directrix  by  comparison  with  (II),  p.  178,  and  transform  to  the  old  coordinates. 

3.  If  Pi  and  P2  are  two  points  whose  polars  with  respect  to  a  circle  Ci  are 

ii  and  L2,  then  —  =  — ,  where  Zi  and  I2  are  the  distances  from  the  center  of 
di      di 

C\  to  Pi  and  P2,  d\  is  the  distance  from  L^  to  Pi,  and  ^2  from  Li  to  P2. 

Hint.  The  center  of  C^  may  be  taken  as  the  origin.  Apply  (IV),  p.  31,  and  the  Rule, 
p.  106. 

4.  Prove  Theorem  XII  and  problem  2  by  means  of  problem  3  and  the  defi- 
nition of  a  conic  (p.  173). 

Hint.  Let  P^  of  problem  3  be  the  center  of  C. 

6.  The  angles  which  two  lines  L\  and  L^.  (Fig.,  p.  311),  which  are  tan- 
gent to  a  circle  (7,  make  with  the  polar  L  of  their  point  of  intersection 
are  evidently  equal.  If  we  reciprocate  the  figure  with  respect  to  a  circle 
Ci,  what  will  be  the  corresponding  theorem  in  the  new  figure  ? 

Hint.  The  polar  reciprocal  of  C  is  a  conic  whose  focus  is  the  center  of  C^  (problem  2). 
To  i,  and  L^  correspond  two  points  on  the  conic,  and  to  their  points  of  contact  correspond 
the  tangents  to  the  conic  at  these  points.  To  L  corresponds  the  point  of  intersection  of 
these  tangents.  Draw  lines  from  the  focus  to  the  points  of  contact  of  the  tangents  and 
to  their  point  of  intersection,  and  apply  the  Corollary  to  Theorem  II,  p.  310. 

Ans.  If  two  tangents  be  drawn  to  a  conic,  the  line  joining  the  focus  to 
their  point  of  intersection  bisects  the  angle  between  the  focal  radii  drawn  to 
the  point  of  contact. 


POLES  AND  POLARS  323 

6,  Obtain  the  following  theorems  in  the  right-hand  column  from  those  in 
the  left-hand  by  means  of  a  polar  reciprocation  with  respect  to  a  circle. 

(a)  Any  tangent  to  a  circle  is  per-  The  lines  from  a  focus  to  any  point 
pendicular  to  the  radius  drawn  to  the  on  a  conic  and  to  the  point  where  the 
point  of  contact.                                            tangent  at  that  point  meets  the  directrix 

are  perpendicular. 

(b)  The  angle  formed  by  two  tan-  The  angle  formed  by  the  focal  radii 
gents  to  a  circle  is  bisected  by  the  line       of  a  conic  drawn  to  its  points  of  inter- 
drawn  from  the  center  to  their  point       'Section  with  any  line  is  bisected  by  the 
of  intersection.                                                line  joining  the  focus  to  the  intersec- 
tion of  that  line  and  the  directrix. 

(c)  The  points  of  intersection  of  Chords  of  a  conic  which  subtend 
tangents  to  a  circle  which  intersect  at  equal  angles  at  the  focus  are  tangent 
a  constant  angle  lie  on  a  concentric  to  a  conic  with  the  same  focus  and 
circle.                                                                directrix. 

137.  Correlations.  Any  transformation  which  makes  the  points  of  one 
figure  correspond  to  the  lines  of  a  second  figure  is  called  a  correlation.  Polar 
reciprocations  with  respect  to  conies  are  the  most  important  correlations. 

A  correlation  is  completely  determined  when  we  are  able  to  find 

1.  The  equation  of  the  line  corresponding  to  a  given  point. 

2.  The  coordinates  of  the  point  corresponding  to  a  given  line. 

We  shall  now  see  that  a  correlation  is  defined  by  an  equation  of  the  form 

(1)  {aiXi  -f  biyi  +  Ci)  X  +  (a2Xi  +  622/1  +  02)2/+  {a^Xi  +  632/1  +  Cs)  =  0, 
which  is  of  the  first  degree  in  x  and  y  and  in  Xi  and  yi. 

The  locus  of  (1)  is  the  line  corresponding  to  a  given  point  Pi{xi,  2/1). 
To  find  the  point  corresponding  to  a  given  line 

(2)  Ax  +  By  +  C  =  0, 

we  suppose  that  Pi  (cci,  yi)  is  the  required  point.     The  equation  of  the  line 
jorresponding  to  Pi  is  (1).     Hence  (1)  and  (2)  have  the  same  locus  and 
lerefore 

aiXi  +  6iyi  +  ci  _  ^23:1  +  62^1  +  C2  _  «3a^i  +  63^1  +  Cs 
'  Z  ~  B  ~  C  ' 

These  equations  may,  in  general,  be  solved  for  Xi  and  2/1. 

As  far  as  defining  the  line  corresponding  to  a  given  point  is  concerned,  the  parenthe- 

in  (1)  might  be  any  complicated  expressions  in  x^  and  y^.    But  if  the  expressions  in 

lose  parentheses  were  not  of  the  first  degree,  then  the  equations  (3)  would  have  more 

lan  one  pair  of  solutions  for  x^  and  y^,  and  hence  there  would  be  more  than  one  point 

corresponding  to  a  given  line. 

In  general  the  point  Pi  will  not  lie  upon  the  locus  of  (1).  The  condition 
that  Pi  should  lie  on  the  locus  of  (1)  is  (Corollary,  p.  63) 

(aiXi  +  6i?/i  +  ci)  xi  +  {a2Xi  +  622/1  +  Cg)  2/1  +  (as^i  +  632/1  +  a^)  =  0, 
or       aiXi2  +  (61  +  02)  xiyi  +  622/1^  +  (ci  +  as)  Xi  +  (C2  +  63)  yi  +  as  =  0. 


324  ANALYTIC  GEOMETRY 

This  is  also  the  condition  that  Pi  shall  lie  upon  the  locus  of  the  equation 

(4)  aix2  +  (61  +  a^)  xy  +  &22/^  +  (ci  +  as)  x  +  (cg  +  63)  y  +  as  =  0. 

The  manner  in  which  the  conic  sections  enter  into  the  theory  of  correlations 
is  thus  given  by 

Theorem  XIII.  The  locus  of  the  points  which  lie  upon  the  lines  corresponding 
to  them  in  the  correlation  defined  by  (1)  is  the  conic  or  degenerate  conic  whose 
equation  is  (4).     • 

It  should  be  noticed  that  the  correlation  defined  by  (1)  is  not,  in  general, 
a  polar  reciprocation  in  the  curve  (4),  for  (1)  is  not  the  equation  of  the  polar 
of  Pi(xi,  yi)  with  respect  to  (4). 

Suppose,  however,  that  61  =  a^,  Ci  =  as,  and  Cg  =  &3-     Then  (4)  becomes 

(5)  ttix^  +  2  a^xy  +  hy^  +  2  asx  +  2  b^y  +  as  =  0, 
and  (1)  becomes 

(aiXi  +  azyi  +  az)x-\-  (a2Xi  +  622/1  +  63)  2/  +  (as^i  +  632/1  +  C3)  =  0, 
or 

(6)  aixix  +  a2  {yiX  +  Xiy)  +  622/12/  +  as  (x  +  Xi)  +  63  (y  +  2/1)  +  C3  =  0. 

The  locus  of  (6)  is  the  polar  of  Pi(Xi,  2/1)  with  respect  to  (5).  Hence  we 
have 

Theorem  XIV.  Ifbi  =  ao,  Ci  =  as,  and  c^  =  b^,  then  the  correlation  defined 
by  (1)  is  a  polar  reciprocation  with  respect  to  the  locus  of  (5). 


CHAPTER    XYI 

CARTESIAN  COORDINATES  IN  SPACE 

138.  Cartesian  coordinates.  The  foundation  of  Plane  Analytic 
Geometry  lies  in  the  possibility  of  determining  a  point  in  the 
plane  by  a  pair  of  real  numbers  (x,  y)  (p.  25).  The  study  of 
Solid  Analytic  Geometry  is  based  on  the  determination  of  a  point 
in  space  by  a  set  of  three  real  numbers  x,  y,  and  z.  This  deter- 
mination is  accomplished  as  follows  : 

Let  there  be  given  three  mutually  perpendicular  planes  inter- 
secting in  the  lines  XX\  YY\  and  ZZ'  which  will  also  be  mutually 
perpendicular.  These  three 
planes  are  called  the  coordinate 
planes  and  may  be  distin- 
guished as  the  ZF-plane,  the 
FZ-plane,  and  the  ZX-plane. 
Their  lines  of  intersection 
are  called  the  axes  of  coordi- 
nates, and  the  positive  direc- 
tions on  them  are  indicated 
by  the  arrowheads.*  The 
point  of  intersection  of  the 
coordinate  planes  is  called 
the  origin. 

Let  P  be  any  point  in  space  and  let  three  planes  be  drawn 
through  P  parallel  to  the  coordinate  planes  and  cutting  the  axes 
2it  A,  B,  and  C.  Then  the  three  numbers  OA  =>:  x,  OB  =  y,  and 
OC  =  z  are  called  the  rectangular  coordinates  of  P. 


*  XX'  and  ZZ'  are  supposed  to  be  in  the  plane  of  the  paper,  the  positive  direction  on 
XX'  being  to  the  right,  that  on  ZZ'  being  upward.  YY'  is  supposed  to  be  perpendicular 
to  the  plane  of  the  paper,  the  positive  direction  being  in  front  of  the  paper,  that  is,  from 
the  plane  of  the  paper  toward  the  reader. 

325 


326 


ANALYTIC  GEOMETRY 


Any  point  P  in  space  determines  three  numbers,  the  coordinates 
of  P.  Conversely,  given  any  three  real  numbers  x,  y,  and  z,  a 
point  P  in  space  may  always  be  constructed  whose  coordinates 
are  x,  y,  and  z.  For  if  we  lay  off  OA  =  x,  OB  =  y,  and  OC  =  z, 
and  draw  planes  through  A,  B,  and  C  parallel  to  the  coordinate 
planes,  they  will  intersect  in  such  a  point  P.     Hence 

Every  point  determines  three  real  numbers,  and  conversely,  three 
real  numbers  determine  a  point. 

The  coordinates  of  P  are  written  (x,  y,  z),  and  the  symbol 
P(x,  y,  z)  is  to  be  read,  "The  point  P  whose  coordinates  are 
x,  y,  and  zJ' 

The  coordinate  planes  divide  all  space  into  eight  parts  called 
octants,  designated  hj.O-XYZ,  0-X'YZ,  etc.  The  signs  of  the 
coordinates  of  a  point  in  any  octant  may  be  determined  by  the 

Rule  for  signs. 

X  is  positive  or  negative  according  as  P  lies  to  the  right  or  left 
of  the  YZ-plane. 

y  is  positive  or  negative  according  as  P  lies  in  front  or  in  back 

of  the  ZX-plane. 

z  is  positive  or  negative  according  as 
P  lies  above  or  below  the  XY-plane. 

If  the  coordinate  planes  are  not 
mutually  perpendicular,  we  still  have 
an  analogous  system  of  coordinates 
called   oblique   coordinates.      In    this 
system   the  coordinates  of  a   point 
^      are   its   distances   from   the  coordi- 
^  nate  planes  measured  parallel  to  the 
axes  instead  of  perpendicular  to  the 
planes.     We  shall  confine  ourselves 
to  the  use  of  rectangular  coordinates. 

Points  in  space  may  be  conveniently  plotted  by  marking  the  same  scale  on  XX' 
and  ZZ'  and  a  somewhat  smaller  scale  on  YY\  Then  to  plot  any  point,  for  example 
(7,  6, 10),  we  lay  ofi  OA  —  1  on  OX,  draw  ylQ  parallel  to  OF  and  equal  to  6  units 
on  0  r,  and  QP  parallel  to  OZ  and  equal  to  10  units  on  OZ. 


CARTESIAN  COORDINATES  IN  SPACE  327 

PROBLEMS 

1.  What  are  the  coordinates  of  the  origin  ? 

2.  Plot  the  following  sets  of  points. 

(a)  (8,  0,2),  (-3,  4,  7),  (0,0,  5). 

(b)  (4,  -3,6),  (-4,  6,0),  (0,8,  0). 

(c)  (10,  3,  -4),  (-4,  0,0),  (0,8,4). 

(d)  (3,  -4,  -8),  (-5,  -6,4),  (8,6,  0). 

(e)  (-4,  -8,  -6),  (3,  0,7),  (6,  -4,2). 

(f)  (-6,4,  -4),  (0,-4,  6),  (9,  7,  -2). 

3.  Where  can  a  point  move  ifcc  =  0?  if  y  =  0?  if  z  =  0? 

4.  Where  can  a  point  move  if  ic  =  0  and  y  =  0?  it  y  =  0  and  z  =  0 ? 
if  2;  =  0  and  X  =  0  ? 

5.  Show  that  the  points  {x,  y,  z)  and  (—  x,  y,  z)  are  symmetrical  with 
respect  to  the  FZ-plane ;  (x,  y,  z)  and  (x,  —  y,  z)  with  respect  to  the  ZX- 
plane ;   (x,  y,  z)  and  ^,  y,  —  z)  with  respect  to  the  XZ-plane. 

6.  Show  that  the  points  (x,  y,  z)  and  (—  x,  -  y,  z)  are  symmetrical  with 
respect  to  ZZ' ;  (x,  y,  z)  and  (x,  —  y,  —z)  with  respect  to  XX' ;  (x,  y,  z)  and 
(-  X,  y,  —z)  with  respect  to  YY' ;  (x,  y,  z)  and  (-  x,  —  y,  -  z)  with  respect 
to  the  origin. 

7.  What  is  the  value  of  z  if  P  (x,  y,  z)  is  in  the  JTF-plane  ?  of  x  if  P  is  in 
the  FZ-plane  ?  of  ?/  if  P  is  in  the  ZA"-plane  ? 

8.  What  are  the  values  of  y  and  z  if  P  (x,  y,  z)  is  on  the  JT-axis  ?  of  z  and 
X  if  P  is  on  the  F-axis  ?  of  x  and  ?/  if  P  is  on  the  Z-axis  ? 

9.  A  rectangular  parallelopiped  lies  in  the  octant  0-XYZ  with  three 
faces  in  the  coordinate  planes.  If  its  dimensions  are  a,  6,  and  c,  what  are 
the  coordinates  of  its  vertices  ? 

139.  Orthogonal  projections.  To  extend  the  first  theorem  of 
projection  (p.  30)  we  define  the  angle  between  two  directed  lines 
in  space  which  do  not  intersect  to  be  the  angle  between  two 
intersecting  directed  lines  (p.  28)  drawn  parallel  to  the  given 
lines  and  having  their  positive  directions  agreeing  with  those  of 
the  given  lines. 

The  definitions  of  the  orthogonal  projection  (p.  29)  of  a  point 
upon  a  line  and  of  a  directed  length  AB  upon  a  directed  line 
^hold  when  the  points  and  lines  lie  in  space  instead  of  in  the 
plane.     It  is  evident  that  the  projection  of  a  point  upon  a  line 


328 


ANALYTIC  GEOMETRY 


may  also  be  regarded  as  the  point  of  intersection  of  the  line  and 
the  plane  passed  through  the  point  perpendicular  to  the  line. 
As  two  parallel  planes  are  equidistant,  then  the  projections  of  a 
directed  length  AB  upon  two  parallel  lines  whose  positive  directions 
agree  are  equal. 

Theorem  I.  First  theorem  of  projection.  If  A  and  B  are  points 
upon  a  directed  line  making  an  angle  of  y  ivith  a  directed  line  CD, 
then  the 

projection  of  the  length  AB  upon  CD  =  AB  cos  y. 

Proof    Draw  CD'  through   A 

parallel  to  CD.     Then  by  defini- 

D'  tion  the  angle  between  AB  and 

^    CD'  equals  y.     Since  CD'  and  AB 

^     intersect  we  may  apply  the  first 

theorem  of  projection  in  the  plane 

(p.  30),  and  hence  the 

projection  of  the  length  AB  upon  C'D'  =  AB  cos  y. 

Since  the  projection  of  AB  on  CD  equals  the  projection  of  AB 
upon  CD'  we  get  (I).  q.e.d. 

Theorem  II.  Second  theorem  of  projection.  If  each  segment  of  a 
broken  line  in  space  he  given  the  direction  determined  in  passing 
continuously  from  one  extremity  to  the  other,  then  the  algebraic 
sum  of  the  projections  of  the  segments  upon  any  directed  line  equals 
the  projection  of  the  closing  line. 

The  proof  given  on  p.  48  holds  whether  the  broken  line  lies  in  the  plane  or  in 
space. 


Corollary  I.  The  projections  on  the  axes  of  coordinates  of  the 
line  joining  the  origin  to  any  point  P  are  respectively  the  coordi- 
nates of  P. 

For  the  projection  of  OP  (Fig.,  p.  325)  upon  OX  equals  the  sum  of  the  projec- 
tions of  OA,  AQ,  and  QP,  which  are  respectively  equal  to  x,  0,  and  0  [by  (I)]. 
Similarly  for  the  projections  on  OTand  OZ. 


CARTESIAN  COORDINATES  IN  SPACE  329 

Corollary  II.  Given  any  two  points  P^  {x^,  y^,  z^)  and  P^  {x^,  2/2?  ^i)i 
then 

a?2  —  a?!  =  projection  of  JP^P^,  upon  XX', 
y^  —  y^  =  projection  of  ^Pi-Pg  upon  YY^, 
z^  —  z^z=  projection  of  -Pi  1*2  upon  ZZK 

For  if  we  project  P\OP<i  and  P\Pi  upon  XX\  we  have  the 

proj.  of  PiO  +  proj.  of  OPi  =  proj.  of  PxP%. 
But  by  Corollary  I, 

proj.  of  P\0  —  —  cci,    proj.  of  OP2  =  'X,^. 
.'.  a;2  —  a;i  =  proj.  of  P1P2  upon  XX\ 
In  like  manner  the  other  formulas  are  proved. 

Corollary  III.  If  the  sides  of  a  polygon  be  given  the  direction 
established  by  passing  continuously  around  the  perimeter,  the  sum 
of  the  projections  of  the  sides  tipon  any  directed  line  is  zero. 

PROBLEMS 

1.  Find  the  projections  upon  each  of  the  axes  of  the  sides  of  the  triangles 
whose  vertices  are  the  following  points  and  verify  the  results  by  Corollary  III. 

(a)  (-  3,  4,  -  8),  (5,  -  6,  4),  (8,  6,  0). 

(b)  (-4,  -8,  -6),  (3,  0,7),  (6,  4,  -2). 

(c)  (10,3,  -4),  (-4,  0,2),  (0,8,  4). 

(d)  (-6,4,-4),  (0,  -4,6),  (9,7,-2). 

2.  If  the  projections  of  P1P2  on  the  axes  are  respectively  3,  —  2,  and  7, 
and  if  the  coordinates  of  Pi  are  (—4,  3,  2),  find  the  coordinates  of  P2. 

Ans.    (-1,  1,  9). 

3.  A  broken  line  joins  continuously  the  points  (6,  0,  0),  (0,  4,  3),  (—  4,  0,  0), 
and  (0,  0,  8).  Find  the  sum  of  the  projections  of  the  segments  and  the  pro- 
jection of  the  closing  line  on  (a)  the  X-axis,  (b)  the  F-axis,  (c)  the  Z-axis, 
and  verify  the  results  by  Theorem  II.     Construct  the  figure. 

4.  A  broken  line  joins  continuously  the  points  (6,  8,  —  3),  (0,  0,  —  3), 
(0,  0,  6),  (-8,  0,  2),  and  (-  8,  4,  0).  Find  the  sum  of  the  projections  of 
the  segments  and  the  projection  of  the  closing  line  on  (a)  the  X-axis,  (b)  the 
F-axis,  (c)  the  Z-axis,  and  verify  the  results  by  Theorem  II.  Construct  the 
figure. 

5.  Find  the  projections  on  the  axes  of  the  line  joining  the  origin  to  each 
of  the  points  in  problem  1. 


330 


ANALYTIC  GEOMETRY 


6.  Find  the  angles  between  the  axes  and  the  line  drawn  from  the  origin  to 


(a)  the  point  (8,  6,  0). 

(b)  the  point  (2,  - 1,  -  2). 


Ans.    cos-^ 


Ans.    cos-i-»  cos 
3 


3      TT 

6'  2 


-i-l>-i-l) 


7.  Find  two  expressions  for  the  projections  upon  the  axes  of  the  line 
drawn  from  the  origin  to  the  point  P{x,  y,  z)  if  the  length  of  the  line  is 
p  and  the  angles  between  the  line  and  the  axes  are  a,  /3,  and  y. 

8.  Find  the  projections  of  the  coordinates  of  P(x,  y,  z)  upon  the  line 
drawn  from  the  origin  to  P  if  the  angles  between  that  line  and  the  axes 
are  a,  j3,  and  y.  Ans.   x  cos  a,  y  cos  j3,  z  cos  7. 

140.  Direction  cosines  ^of   a   line.    The   angles  a,  /3,  and  y 

between  a  directed  line  and  the  axes  of  coordinates  are  called 

the  direction  angles  of  the  line. 

If  the  line  does  not  intersect  the  axes,  then  by  definition  (p.  327)  a,  j3,  and  y 
are  the  angles  between  the  axes  and  a  line  drawn  through  the  origin  parallel  to 
the  given  line  and  agreeing  with  it  in  direction. 

The  cosines  of  the  direction  angles  of  a  line  are  called  the 
direction  cosines  of  the  line. 

Reversing  the  direction  of  a  line  changes  the  signs  of  the  direc- 
tion cosines  of  the  line. 

For  reversing  the  direction  of  a  line  changes  a,  /3,  and  7  into  (p.  28)  it  —  a, 
TT  —  /3,  and  it  —  y  respectively,  and  (5,  p.  20)  cos  (jt  —  x)  =  —  cos  x. 

Theorem  III.    If  a.,  ft,  and  y  are  the  direction  angles  of  a  line,  then 


(III) 


cos'*  a  +  cos^^  +  cos^*  7  =  1- 

That  is,  the  sum  of  the  squares  of  the 
direction  cosines  of  a  line  is  unity. 

Proof.  Let  ^S  be  a  line  whose 
direction  angles  are  a,  ji,  and  y. 
Through  0  draw  OP  parallel  to 
AB  and  let  OP  =  p.  By  definition 
(p.  327)  Z  XOP  =  a,  Z  Y0P  =  (3, 
ZZOP=y.  Projecting  OP  on  the 
axes,  we  get  by  Corollary  I,  p.  328, 
and  Theorem  I,  p.  328, 

p  cos  (3,  z  =  p  cos  y. 


CARTESIAN  COORDINATES  IN  SPACE  331 

Projecting  OP  and  OCQP  on  OP,  we  get  (Theorems  I  and  II) 
(2)  p  =  xcosa  -^  y  cosft  -\-  z  cos  y. 

Substituting  from    (1)   in   (2)   and  dividing  by  p,  we  obtain 

(HI).  Q.E.D. 

^.  cosa      COS/8      cosy 
Corollary,    if =  — r-^  = }  then 

a  '  b 

cos  a  = J     cos  £j  = 


±  Va2  +  62  _^  c^  ±  Va^  +  &2  +  c2 

c 

COS  y  = ^ 

That  is,  if  the  direction  cosines  of  a  line  are  proportional  to  three 
numbers,  they  are  respectively  equal  to  these  numbers  each  divided 
by  the  square  root  of  the  sum  of  their  squares. 

For  if  r  denotes  the  common  value  of  the  given  ratios,  then 
(3)  cos  a  =  ar,    cos  j3  =  br,    cos  y  =  cr. 

Squaring,  adding,  and  applying  (III) , 

I=:r2(a2  +  62  +  02). 


±  Va2  +  62  _|.  c2 
Substituting  in  (3),  we  get  the  values  of  cos  a,  cos  /3,  and  cos  y  to  be  derived.  ' 

If  a  line  cuts  the  XF-plane,  it  will  be  directed  upward  or  downward  according 
as  cos  7  is  positive  or  negative. 

If  a  line  is  parallel  to  the  XF-plane,  cos  7  —  0  and  it  will  be  directed  in  front 
or  in  hack  of  the  ZX-plane  according  as  cos  j8  \&  positive  or  negative. 

If  a  line  is  parallel  to  the  X-axis,  cos  /S  =  cos  7  =  0,  and  its  positive  direction 
will  agree  or  disagree  with  that  of  the  X-axis  according  as  cos  or  =  1  or  —  1 . 

These  considerations  enable  us  to  choose  the  sign  of  the  radical  in  the  Corollary 
so  that  the  positive  direction  on  the  line  shall  be  that  given  in  advance. 

141.  Lengths. 

Theorem  IV.  The  length  I  of  the  line  joining  two  points 
Pi(xi,  1/1,  ^1)  «^^  A (^2,  1/2,  ^2)  ^'-^  ff^^en  by 


(IV)         I  =  V(^,  -  ^.,y  +  (y,  -  y,y  +.  (z,  -  z,y. 


332 


ANALYTIC   GEOMETRY 


Proof.    Let  the  direction  angles  of  the  line  PiPg  he  a,  ^,  and  y. 
Projecting  P^P^  on  the  axes,  we  get,  by  Theorem  I,  p.  328, 
and  Corollary  II,  p.  329, 

(1)  I  cos  a  =  X2  —  Xi,    I  COS  /3  =  1/2  —  ^Jl)    I  cos  y  =  ^2  —  ^1- 

Squaring  and  adding, 
Z^  (cos^  a  +  cos^^  /?  +  cos^  y)  =  {x^  —  x^y-\-  (2/2  —  ViY  +  («2  —  ^l)^ 

=  (^1  -  ^lY  +  (i/i  -  2/2)'  +  («i  -  ^2)'. 


Applying  (III),  p.  330,  and  taking  the  square  root,  we  get  (IV). 

Q.E.D. 

Corollary.    The  dh'ection  cosines  of  the  line  drawn  from  P^  to 
P2  are  proportional  to  the  projections  of  P1P2  on  the  axes. 

cos  a   _    cos  /3 


For,  from  (1), 


X2  —  Xl 


-2/1 


cos  7 

^2  -  21 ' 


since  each  ratio  equals  -y  and  the  denominators  are  the  projections  of  P1P2  on 

the  axes  (Corollary  II,  p,  329). 


^t 


V 


rj;E=n'-i 


/I  ~ 


1/ 


V 


If  we  construct  a  rectangular  parallelepiped 
by  passing  planes  through  P^  and  P2  parallel 
to  the  coordinate  planes,  its  edges  will  be  paral- 
lel to  the  axes  and  equal  numerically  to  the 
projections  of  P1P2  upon  the  axes.  P1P2  will 
be  a  diagonal  of  this  parallelepiped,  and  hence 
/  ^  l^  will  equal  the  sum  of  the  squares  of  its 
three  dimensions.  AVe  have  thus  a  second 
method  of  deriving  (IV). 


PROBLEMS 


1.  Find  the  length  and  the  direction  cosines  of  the  line  drawn  from 

(a)  Pi  (4,  3,  -2)  to  P2(-2,  1,  —5). 

(b)  Pi  (4,  7,  -2)  to  P2(3,  5,  -4). 

(c)  Pi  (3,  -  8,  6)  to  P2  (6,  -  4,  6). 


Ans.  7,  -  f,  -  f,  -  f. 
Ans.  3,  -i,  -  I,  -  f. 
Ans.    5.  f,  f,  0. 


2.  Find  the  direction  cosines  of  a  line  directed  upward  if  they  are  propor- 
tional to  (a)  3,  6,  and  2  ;  (b)  2,  1,  and  -  4 ;  (c)  1,-2,  and  3. 

Ans.   (a)  f,  f,  f;   (b)       ^  ^ 


-2 


^ ^      •(c)J-, , 

V21' -V2T' +  V21'       V14' VTi' Vi4 


CARTESIAN  COORDmATES  IN  SPACE  333 

3.  Find  the  lengths  and  direction  cosines  of  the  sides  of  the  triangles 
whose  vertices  are  the  following  points ;  then  find  the  projections  of  the  sides 
upon  the  axes  by  Theorem  I,  p.  328,  and  verify  by  Corollary  III,  p.  329. 

(a)  (0,  0,  3),  (4,  0,  0),  (8,  0,  0). 

(b)  (3,2,0),  (-2,5,7),  (1,  -3,  -5). 

(c)  (-4,0,  6),  (8,  2,-1),  (2,  4,  6). 

(d)  (3,  -  3,  -  3),  (4,  2,  7),  (-  1,  -  2,  -  5). 

4.  In  what  octant  {0-XYZ,  0-X'YZ,  etc)  will  the  positive  part  of 
a  line  through  0  lie  if 

(a)  cos  a  >  0,  cos  ^3  >  0,  cos  7  >  0  ?  (e)  cos  a  <  0,  cos  j8  >  0,  cos  7  >  0  ? 

(b)  cosa>0,  cos/3>0,  cos7<0?  (f)  cosa<0,  cos/3<0,  cos7>0? 

(c)  cos  a:  >  0,  cos  j3  <  0,  cos  7  <  0  ?  (g)  cos  a  <  0,  cos  /3  <  0,  cos  7  <  0  ? 

(d)  cos  a  >  0,  cos  /3  <  0,  cos  7  >  0  ?  (h)  cos  or  <  0,  cos  /3  >  0,  cos  7  <  0  ? 

5.  What  is  the  direction  of  a  line  if  cos  a  =  0?  cosjQ  =  0  ?  cos  7  =  0? 
cos  a  =  cos  /3  =  0  ?  cos  /3  =  cos  7  =  0?  cos  7  =  cos  ex  =  0? 

6.  Find  the  projection  of  the  line  drawn  from  the  origin  to  Pi  (5,  —  7,  6) 
upon  a  line  whose  direction  cosines  are  f,  —  7,  and  f.  Ans.    9. 

Hint.  The  projection  of  OP^  on  any  line  equals  the  projection  of  a  broken  line  whose 
segments  equal  the  coordinates  of  P^. 

7.  Find  the  projection  of  the  line  drawn  from  the  origin  to  Pi  (xi,  ?/i,  Zi) 
upon  a  line  whose  direction  angles  are  a,  p,  and  7. 

Ans.    Xi  cos  a  +  yi  cos  ^  +  Zi  cos  7. 

8.  Show  that  the  points  (-  3,  2,  -  7),  (2,  2,  -  3),  and  (-  3,  6,  -  2)  are 
the  vertices  of  an  isosceles  triangle. 

9.  Show  that  the  points  (4,  3,  -  4),  (-  2,  9,  -  4),  and  (-  2,  3,  2)  are  the 
vertices  of  an  equilateral  triangle. 

10.  Show  that  the  points  (-4,  0,  2),  (-1,  3  V3,  2),  (2,  0,  2),  and 
(-1,  V3,  2  +  2  Vq)  are  the  vertices  of  a  regular  tetraedron. 

11.  What  does  formula  (IV)  become  if  Pi  and  P2  lie  in  the  XF-plane  ? 
in  a  plane  parallel  to  the  XF-plane  ? 

12.  Show  that  the  direction  cosines  of  the  lines  joining  each  of  the  points 
(4,  -  8,  6)  and  (-  2,  4,  -  3)  to  the  point  (12,  -  24,  18)  are  the  same.  How 
are  the  three  points  situated  ? 

13.  Show  by  means  of  direction  cosines  that  the  three  points  (3,  —  2,  7), 
(6,  4,  —  2),  and  (5,  2,  1)  lie  on  a  straight  line. 

14.  What  are  the  direction  cosines  of  a  line  parallel  to  the  X-axis  ?  to  the 
F-axis  ?  to  the  Z-axis  ? 


334 


ANALYTIC  GEOMETRY 


15.  What  is  the  value  of  one  of  the  direction  cosines  of  a  line  parallel 
to  the  XY-plane  ?  the  YZ-plane  ?  the  ZX-plane  ?  What  relation  exists 
between  the  other  two  ? 

16.  Show  that  the  point  (—1,  —  2,  —  1)  is  on  the  line  joining  the  points 
(4,  —  7,  3)  and  (—  6,  3,  —  5)  and  is  equally  distant  from  them. 

7t  It 

17.  If  two  of  the  direction  angles  of  a  line  are  —  and  - ,  what  is  the  third  ? 

3  *    ,         TT       27r 

Ans.    —or 

3        3 

18.  Find  the  direction  angles  of  a  line  which  is  equally  inclined  to  the 
three  coordinate  axes.  Ans.    a  =  p  =  y  =  cos-i ^^S. 

19.  Find  the  length  of  a  line  whose  projections  on  the  axes  are  respectively 

(a)  6,  -  3,  and  2.  Ans.    7. 

(b)  12,  4,  and  -  3.  Ans.    13. 

(c)  -  2,  -  1,  and  2.  Ans.    3. 

142.  Angle  between  two  directed  lines. 

Theorem  V.  If  a,  ^,  y  and  a',  /3',  y'  are  the  direction  angles  of 
two  directed  lines,  then  the  angle  6  between  them  is  given  by 

(V)  cos  ^  =  cos  a  cos  a'  +  cos  p  cos  jff'  +  cos  y  cos  y'. 

Proof  Draw  OP  and  OP' 
parallel  to  the  given  lines  and 
let  OP  ==  p.  Then  by  definition, 
p.  327, 

Z  POP'  =  6. 

Project  OP  and  OABP  on  OP'. 
^  Then  by  Theorem  I,  p.  328,  and 
Theorem  II,  p.  328, 

(1)  p  cos  6 

=  X  cos  a'-\-  y  cos  ^'+  ^  COS  y'. 

Projecting  OP  on  the  axes  (Corollary  I,  p.  328,  and  Theorem  I), 
(2).  X  =  p  cos  a,     y  =  p  COS  ^,     z  =  p  cos  y. 

Substituting  in  (1)  from  (2)  and  dividing  by  p,  we  obtain  (V). 

Q.E.D. 


'AX 


CARTESIAN  COORDINATES  IN  SPACE  336 

Theorem  VI.  If  a,  ^,  y  and  a',  p\  y'  are  the  direction  angles  of 
two  lines,  then  the  lines  are 

(a)  parallel  and  in  the  same  direction*  when  and  only  when 

a=a',/3  =  /3',y  =  y'', 

(b)  perpendicular  "f  ivhen  and  only  ivhen 

cos  a  cos  a'-\-  COS  P  COS  /8'  H-  cos  y  COS  y'  =  0. 

That  is,  two  lines  are  parallel  and  in  the  same  direction  when 
and  only  luhen  their  direction  angles  are  equal,  and  perpendicular 
when  and  only  when  the  sum  of  the  products  of  their  direction 
cosines  is  zero. 

Proof  The  condition  for  parallelism  follows  from  the  fact 
that  both  lines  will  be  parallel  to  and  agree  in  direction  with  the 
same  line  through  the  origin  when  and  only  when  their  direction 
angles  are  equal. 

The  condition  for  perpendicularity  follows  from  (Y),  for  if 

IT 

0  =  —}  then  cos  ^  ==  0,  and  conversely.  q.e.d. 

Li 

Corollary.  If  the  direction  cosines  of  the  lines  are  proportional 
to  a,  h,  c  and  a\  b',  c',  then  the  conditions  for  parallelism  and 
perpendicularity  are  respectively 

^  b  C  ,  T  7  I  t  r. 

—  =  -  =  -,      aa' +  bb' -\- cc' =  0. 
a'      b       c' 

143.  Point  of  division. 

Theorem  VII.  The  coordinates  (x,  y,  z)  of  the  point  of  division 
P  on  the  line  joining  Pi(xi,  yi,  Zi)  and  P2(^2>  y2}  ^2)  such  that  the 
ratio  of  the  segments  ^s 

are  given  by  the  formulas 

(VII)   «=  =  ^Vtx''  ^  =  ^ttx''  "  =  ttx- 

This  is  proved  as  on  p.  39. 

*  They  will  he  parallel  and  differ  in  direction  when  and  only  when  the  direction 
angles  are  supplementary. 

t  Two  lines  in  space  are  said  to  he  perpendicular  when  the  angle  hetween  them  is  — » 
but  the  lines  do  not  necessarily  intersect. 


336  ANALYTIC  GEOMETRY 

Corollary.    The  coordinates  (x,  y,  z)  of  the  middle  point  P  of  the 
line  joining  P\{xi,  yi,  z^)  and  ^2(^2?  1/21  ^2)  ^^^ 

03  =  i(Xi  +  a?2),     2/  =  i(2/i  +  Vi),     «  =  1(^1  +  «2)-   • 

PROBLEMS 

1.  Find   the    angle   between    two   lines   whose    direction   cosines  are 
respectively 

(a)  f,  f,  -f  and  a,  -  |,  f  Ans.   |. 

(b)  f,  -  1,  I  and  -  T?3,  r%,  {h  ^ns.    cos-iif. 

(c)  f,  -  f,  1  and  f,  f,  f.  ^ns.    cos-i(-  f^). 


3.  Show  that  the  lines  joining  the  following  pairs  of  points  are  either 
parallel  or  perpendicular,  -^ 

(a)  (3,  2,  7),  (1,  4,  6)    and  (7,  -  5,  9),  (5,  -  3,  8). 

(b)  (13,  4,  9),  (1,  7,  13)    and  (7,  16,  -  6),  (3,  4,  -  9). 

(c)  (-  6,  4,  -  3),  (1,  2,  7)  and  (8,  -  6,  10),  (15,  -  7,  20). 

4.  Find  the  coordinates  of  the  point  dividing  the  line  joining  the  follow- 
ing points  in  the  ratio  given. 

(a)  (3,  4,  2),  (7,  -  6,  4),  \  =  \.  Ans.    (-V-,  |,  |). 

(b)  (-  1,  4,  -  6),  (2,  3,  -  7),  X  =3  -  3.  Ans.    (|,  |,  -  ^i). 

(c)  (8,  4,  2),  (3,  9,  6),  \  =  -  i.  Ans.    (V,  f ,  0). 

(d)  (7,  3,  9),  (2,  1,  2),  X  =  4.  Ans.    (3,  |,  -U). 

5.  Show  that  the  points  (7,  3,  4),   (1,  0,  6),  and  (4,  5,   —  2)  are  the 
vertices  of  a  right  triangle. 

6.  Showthatthepoints(-6,  3,  2),  (3,  -  2,  4),  (5,  7,  3),  and  (- 13, 17,-1) 
are  the  vertices  of  a  trapezoid. 

7.  Show  that  the  points  (3,  7,  2),  (4,  3,  1),  (1,  6,  3),  and  (2,  2,  2)  are  the 
vertices  of  a  parallelogram. 

8.  Show  that  the  points  (6,  7,  3),  (3,  11,  1),  (0,  3,  4),  and  (-  3,  7,  2)  are 
the  vertices  of  a  rectangle. 

9.  Show  that  the  points  (6,  -  6,  0),  (3,  -  4,  4),  (2,  -  9,  2),  and  (-  1, 
—  7,  6)  are  the  vertices  of  a  rhombus. 

10.  Show  that  the  points  (7,  2,  4),  (4,  -  4,  2),  (9,  -  1, 10),  and  (6,  -  7,  8) 
are  the  vertices  of  a  square. 


CARTESIAN  COORDINATES  IN  SPACE  337 

11.  Show  that  each  of  the  following  sets  of  points  lies  on  a  straight  line, 
and  find  the  ratio  of  the  segments  in  which  the  third  divides  the  line  joining 

the  first  to  the  second. 

(a)  (4,  13,  3),  (3,  6,  4),  and  (2,  -  1,  5).  Ans.    -  2. 

(b)  (4,  -  5,  -  12),  (-  2,  4,  6),  and  (2,  -  2,  -  6).  Ans.    ^ 

(c)  (-  3,  4,  2),  (7,  -  2,  6),  and  (2,  1,  4).  Ans.    1.     ' 

12.  Find  the  lengths  of  the  medians  of  the  triangle  whose  vertices  are 
the  points  (3,  4,  -  2),  (7,  0,  8),  and  (-  6,  4,  6).       Ans.   VTlS,  V89,  2V29. 

13.  Show  that  the  lines  joining  the  middle  points  of  the  opposite  sides  of 
the  quadrilaterals  whose  vertices  are  the  following  points  bisect  each  other. 

(a)  (8,  4,  2),  (0,  2,  5),  (-  3,  2,  4),  and  (8,  0,  -6). 

(b)  (0,  0,  9),  (2,  6,  8),  (-  8,  0,  4),  and  (0,  -  8,  6). 

(c)  Pi(xi,  yi,  zi),  P2{X2,  2/2,  22),  Psixs,  2/3,  Zs),  P^ix^,  2/4,  Z4). 

14.  Show  that  the  lines  joining  successively  the  middle  points  of  the  sides 
of  any  quadrilateral  form  a  parallelogram. 

15.  Find  the  projection  of  the  line  drawn  from  Pi  (3,  2,  —  6)  to  P2(—  3, 
5,  —  4)  upon  a  line  directed  upward  whose  direction  cosines  are  proportional 
to  2,  1,  and  -  2.  Ans.   4i. 

16.  Find  the  projection  of  the  line  djawn  from  Pi  (6,  3,  2)  to  P2(4,  2,  0) 
upon  the  line  drawn  from  P3(7,  -  6/6)  to  P4(-  5,  -  2,  3).         Ans.    ff. 

17.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of  the 
triangle  whose  vertices  are  (3,  6,  —  2),  (7,  —  4,  3),  and  (—  1,  4,  —  7). 

Ans.    (3,  2,  -  2). 

18.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of 
the  triangle  whose  vertices  are  any  three  points  Pi,  P2,  and  P3. 

Ans.   [i  (xi  +  X2  +  X3),  i  {yi  +  2/2  +  2/3),  i  (^i  +  Z2  +  23)]. 

19.  The  three  lines  joining  the  middle  points  of  the  opposite  edges  of  a 
tetraedron  pass  through  the  same  point  and  are  bisected  at  that  point. 

20.  The  four  lines  drawn  from  the  vertices  of_any  tetraedron  to  the  point 
of  intersection  of  the  medians  of  the  opposite  face  meet  in  a  point  which 
is  three  fourths  of  the  distance  from  each  vertex  to  the  opposite  face  (the 
center  of  gravity  of  the  tetraedron). 


CHAPTER   XYII 
SURFACES,  CURVES,  AND  EQUATIONS 

144.  Loci  in  space.  In  Solid  Geometry  it  is  necessary  to  con- 
sider two  kinds  of  loci : 

1.  The  locus  of  a  point  in  space  which  satisfies  one  given  con- 
dition is,  in  general,  a  surface. 

Thus  the  locus  of  a  point  at  a  given  distance  from  a  fixed  point  is  a  sphere, 
and  the  locus  of  a  point  equidistant  from  two  fixed  points  is  the  plane  which  is  , 
perpendicular  to  the  line  joining  the  given  points  at  its  middle  point. 

2.  The  locus  of  a  point  in  space  which  satisfies  two  conditions  * 

is,  in  general,  a  curve.     For  the  locus  of  a  point  which  satisfies 

either  condition  is  a  surface,  and  hence  the  points  which  satisfy 

both  conditions  lie  on  two  surfaces,  that  is,  on  their  curve  of 

intersection. 

Thus  the  locus  of  a  point  which  is  at  a  given  distance  r  from  a  fixed  point  Pi 
and  is  equally  distant  from  two  fixed  points  P^.  and  P3  is  the  circle  in  which  the 
sphere  whose  center  is  Pi  and  whose  radius  is  r  intersects  the  plane  which  is 
perpendicular  to  P2P3  at  its  middle  point. 

These  two  kinds  of  loci  must  be  carefully  distinguished. 

145.  Equation  of  a  surface.     First  fundamental  problem.   If 

any  point  P  which  lies  on  a  given  surface  be  given  the  coordinates 
(x,  y,  z),  then  the  condition  which  defines  the  surface  as  a  locus 
will  lead  to  an  equation  involving  the  variables  x,  y,  and  z. 

The  equation  of  a  surface  is  an  equation  in  the  variables  x,  y, 
and  z  representing  coordinates  such  that : 

1.  The  coordinates  of  every  point  on  the  surface  will  satisfy 
the  equation. 

2.  Every  point  whose  coordinates  satisfy  the  equation  will  lie 
upon  the  surface. 

*  The  number  of  conditions  must  be  counted  carefully.  Thus  if  a  point  is  to  be  equi- 
distant from  three  fixed  points  P^,  P^,  and  P3,  it  satisfies  two  conditions,  namely,  of  being 
equidistant  from  P^  and  P^  and  from  P^  and  P3. 

338 


SURFACES,  CURVES,  AND  EQUATIONS  339 

If  the  surface  is  defined  as  the  locus  of  a  point  satisfying  one 
condition,  its  equation  may  be  found  in  many  cases  by  a  Rule 
analogous  to  that  on  p.  53. 

Ex.  1.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from 
Pi  (3,  0,  -  2)  is  4. 

Solution.  Let  P  {x,  y,  z)  be  any  point  on  the  locus.  The  given  condition 
may  be  written  p  p  _  4 


By  (IV),  p.  331,     PiP  =  V{x  -  3)2  +  y2  +  ^^  4.  2)2. 

.-.  V(x  -  3)2  +  2/2  +  (z  +  2)2  =  4. 
Simplifying,  we  obtain  as  the  required  equation 

x2  +  2/2  +  22  _  6x  +  42;  -  3  =  0. 

That  this  is  indeed  the  equation  of  the  locus  should  be  verified  as  in  Ex.  1, 
52,  and  Ex.  1,  p.  53. 

PROBLEMS 

1 .  Find  the  equation  of  the  locus  of  a  point  which  is 

(a)  3  units  above  the  XF-plane. 

(b)  4  units  to  the  right  of  the  FZ-plane. 

(c)  5  units  below  the  XF-plane. 

(d)  10  units  back  of  the  ZX-plane.-*- 

(e)  7  units  to  the  left  of  the  FZ-plane. 

(f)  2  units  in  front  of  the  ZX-plane. 

2.  Find  the  equation  of  the  plane  which  is  parallel  to 

(a)  the  XF-plane  and  4  units  above  it. 

(b)  the  XF-plane  and  5  units  below  it. 

(c)  the  ZX-plane  and  3  units  in  front  of  it. 

(d)  the  FZ-plane  and  7  units  to  the  left  of  it. 

(e)  the  ZX-plane  and  2  units  back  of  it. 

(f)  the  FZ-plane  and  4  units  to  the  right  of  it. 

3.  Find  the  equation  of  the  sphere  whose  center  is  the  point 

(a)  (3,  0,  4)  and  whose  radius  is  5. 

Ans.   ic2  +  2/2  +  2=2  -  6x  -  8 2;  =  0. 

(b)  (—3,  2,  1)  and  whose  radius  is  4. 

Ans.   ic2  4-2/2  +  z2  +  6x-42/-22;-2=0. 

(c)  (6,  4,  0)  and  whose  radius  is  7. 

Ans.   x2  -f  2/2  +  z2  -  12  X  -  8  2/  -f-  3  =  0. 

(d)  {a,  /3,  7)  and  whose  radius  is  r. 

Ans.    x2  +  y2  +  22  _  2  ax  -  2  /??/  -  2  72;  +  ^2  +  ^2  +  ^2  _  y2  =  0. 


340  ANALYTIC  GEOMETRY 

4.  What  are  the  equations  of  the  coordinate  planes  ? 

5.  What  is  the  form  of  the  equation  of  a  plane  which  is  parallel  to  the 
XY-plane  ?   the  FZ-plane  ?   the  Z X-plane  ? 

6.  Find  the  equation  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points 

(a)  (3,  2,  -  1)  and  (4,  -  3,  0).  '  Ans.  2x -lOy -\- 2 z -11  =  0. 

(b)  (4,  -  3,  6)  and  (2,  -  4,  2).  Ans.  4x  +  2?/  +  82  -  37  =  0. 

(c)  (1,  3,  2)  and  (4,  -  1,  1).  Ans.  Sx  -  4:y  -  z  -  2  =  0. 

(d)  (4,  -  6,  -  8)  and  (- 2,  7,  9).  Ans.  6x  -  13?/ -  172  +  9  =  0. 

7.  Find  the  equations  of  the  six  planes  drawn  through  the  middle  points 
of  the  edges  of  the  tetraedron  whose  vertices  are  the  points  (5,  4,  0), 
(2,  —5,  —4),  (1,  7,  —5),  and  (—4,  3,  4)  which  are  perpendicular  to  the 
edges,  and  show  that  they  all  pass  through  the  point  (—1,  1,  —  2). 

8.  What  are  the  equations  of  the  faces  of  the  rectangular  parallelepiped 
which  has  one  vertex  at  the  origin,  three  edges  lying  along  the  coordinate 
axes,  and  one  vertex  at  the  point  (3,  5,  7)  ? 

9.  Find  the  equation  of  the  sphere  whose  center  is  the  point  (6,  2,  3) 
which  passes  through  the  origin.     Ans.  x^  -]-  y'^  -{■  z^  —  12 x  —  4y  —  6z  =  0. 

10.  Find  the  equation  of  the  locus  of  a  point  which  is  three  times  as  far 
from  the  point  (2,  6,  8)  as  from  (4,  —  2,  4)  and  determine  the  nature  of  the 
locus  by  comparison  with  the  answer  to  problem  3,  (d). 

11.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  the  squares  of 
whose  distances  from  (1,  3,  —  2)  and  (6,  —  4,  2)  is  50  and  determine  the 
nature  of  the  locus  by  comparison  with  the  answer  to  problem  3,  (d). 

146.  Planes  parallel  to  the  coordinate  planes.  We  may  easily 
prove 

Theorem  I.    The  equation  of  a  plane  which  is 

parallel  to  the  XY-plane  has  the  form  z  =  constant; 
parallel  to  the  YZ -plane  has  the  form  x  =  constant; 
parallel  to  the  ZX-plane  has  the  form      y  =  consta7it. 

147.  Equations  of  a  curve.    First  fundamental  problem.   If 

any  point  P  which  lies  on  a  given  curve  be  given  the  coordinates 
{x,  y,  z),  then  the  two  conditions  which  define  the  curve  as  a 
locus  will  lead  to  two  equations  involving  the  variables  x,  y, 
and  z. 


SURFACES,  CURVES,  AND  EQUATIONS 


341 


The  equations  of  a  curve  are  two  equations  in  the  variables 
X,  y,  and  z  representing  coordinates  such  that: 

1.  The  coordinates  of  every  point  on  the  curve  will  satisfy- 
both  equations. 

2.  Every  point  whose  coordinates  satisfy  both  equations  will 
lie  on  the  curve. 

If  the  curve  is  defined  as  the  locus  of  a  point  satisfying  two 
conditions,  the  equations  of  the  surfaces  defined  by  each  condi- 
tion separately  may  be  found  in  many  cases  by  a  Rule  analogous 
to  that  on  p.  53.    These  equations  will  be  the  equations  of  the  curve. 

Ex.  1.  Find  the  equations  of  the  locus  of  a  point  whose  distance  from 
the  origin  is  4  and  which  is  equally  distant  from  the  points  Pi  (8,  0,  0)  and 
A  (0,8,0).  ^^ 

Solution.  First  step.  Let  P{x,  y,  z) 
be  any  point  on  the  locus. 

Second  step.  The  given  conditions  are 
(1)  P0  =  4,     PPi  =  PP2. 

Third  step.    By  (IV),  p.  331, 

Vx2  +  ?/2  +  Z^, 


PO 

PPl  =  V(X  -  8)2  +  2/2  +  z% 

PP2  =  Vx2  +  (y  -  8)2  +  z2. 
Substituting  in  (1),  we  get 


(8,0,0) 


^       Vx2  +  y2  +  2;2  =  4,     V(a;  -  8)2  +  2/2  +  ^2  =  Vx"^  +  {y  -  8)2  +  z^ 
Squaring  and  reducing,  we  have  the  required  equations,  namely, 

x2  +  2/2  +  22  =  16,     x-y  =  0. 
These  equations  should  be  verified  as  in  Ex,  1,  p,  52. 

Ex.  2.  Find  the  equations  of  the  circle  lying  in  the  XF-plane  whose  center 
is  the  origin  and  whose  radius  is  5. 

Solution.    In  Plane  Geometry  the  equation  of  the  circle  is  (Corollary,  p.  68) 

(2)  x^  +  y^  =  25. 

Regarded  as  a  problem  in  Solid  Geometry  we  must  have  two  equations 
which  the  coordinates  of  any  point  P(x,  y,  z)  which  lies  on  the  circle  must 
satisfy.     Since  P  lies  in  the  XF-plane, 

(3)  2  =  0. 

Hence  equations  (2)  and  (3)  together  express  that  the  point  P  lies^m  the 
XF-plane  and  on  the  given  circle.     The  equations  of  the  circle  are  therefore 
x2  +  ?y2  =  25,     z  =  0. 


342  ANALYTIC  GEOMETRY 

The  reasoning  in  Ex.  2  is  general.     Hence 

If  the  equation  of  a  curve  in  the  XY-jplane  is  known,  then  the 
equations  of  that  curve  regarded  as  a  curve  in  space  are  the  given 
equation  and  «  =  0. 

An  analogous  statement  evidently  applies  to  the  equations  of  a 
curve  lying  in  one  of  the  other  coordinate  planes. 

From  Theorem  I,  p.  340,  we  have  at  once 

Theorem  II.    The  equations  of  a  line  ivhich  is  parallel  to 
the  X-axis  have  the  form       y  =  constant,     z  =  constant; 
the  Y-axis  have  the  form       z  =  constant,     x  =  constant; 
the  Z-axis  have  the  form       x  =  constant,     y  =  constant. 

PROBLEMS 

1.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  3  units  above  the  JTF-plane  and  4  units  to  the  right  of  the  FZ-plane. 
•    (b)  5  units  to  the  left  of  the  YZ-plan.e  and  2  units  in  front  of  the  ZX-plane. 

(c)  4  units  back  of  the  ZX-plane  and  7  units  to  the  left  of  the  FZ-plane. 

(d)  9  units  below  the  XY-plane  and  4  units  to  the  right  of  the  rZ-plane. 

2.  Find  the  equations  of  the  straight  line  which  is 

(a)  5  units  above  the  XY-plane  and  2  units  in  front  of  the  ZX-plane. 

(b)  2  units  to  the  left  of  the  YZ-plane  and  8  units  below  the  JTF-plane. 

(c)  3  units  to  the  right  of  the  FZ-plane  and  5  units  from  the  Z-axis. 

(d)  13  units  from  the  X-axis  and  5  units  back  of  the  ZX-plane. 

(e)  parallel  to  the  F-axis  and  passing  through  (3,  7,  —  5). 

(f)  parallel  to  the  Z-axis  and  passing  through  (—4,  7,  6). 

3.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  5  units  above  the  XF-plane  and  3  units  from  (3,  7,  1). 

Ans.    z  =  5,  x^  +  y^  +  z^-Gx-Uy  -2z  + 60  =  0. 

(b)  2  units  from  (3,  7,  6)  and  4  units  from  (2,  5,  4). 

Ans.   x2  -h  ?/2  +  ^2  -  6  X  -  14  i/  -  12  2;  +  90  =  0, 
x2  -f-  2/2  -I-  22  _  4x  _  10 ?/  -  8 2;  +  29  ^  0. 

(c)  5  units  from  the  origin  and  equidistant  from  (3,  7,  2)  and  (  —  3,  -  7,  —2). 

Arts.   x2  -f-  ?/2  -}-  z2  _  25  =  0,  Sx  +  7 y -\- 2z  =  0. 

(d)  equidistant  from  (3,  5,  —  4)  and  (—  7,  1,  6),  and  also  from  (4,  —  6,  3) 
and  (-  2,  8,  5).  Ans.    6x -\- 2y  -  5z  +  11  =0,  3x-7y-z  +  8  =  0. 

(e)  equidistant  from  (2,  3,  7),  (3,  -  4,  6),  and  (4,  3,  -  2). 

Ans.   2x-Uy  -2z-{-l=:0,  x+ly-Sz  +  16  =  0. 


SURFACES,  CURVES,  AND  EQUATIONS  343 

4 .  What  are  the  equations  of  the  edges  of  a  rectangular  parallelepiped  whose 
dimensions  are  a,  b,  and  c,  if  three  of  its  faces  coincide  with  the  coordinate 
planes  and  one  vertex  lies  in  0-XYZ  ?  in  0-XY'Z  ?  in  0-X'Y'Z  ? 

5.  What  are  the  equations  of  the  axes  of  coordinates  ? 

6.  The  following  equations  are  the  equations  of  curves  lying  in  one  of  the 
coordinate  planes.  What  are  the  equations  of  the  same  curves  regarded  as 
curves  in  space? 

(a)  2/2  =  4a;,  (e)  x2  +  42  +  6x  =  0. 

(b)  x^  +  z^  =  16.  (f)  y2-z^-4y  =  0. 

(c)  8x2  -  2/2  =  64.  (g)  yz^-\-z^-Qy  =  0. 

(d)  4  22+ 9  2/2  =  36.  (h)  z2_4a;2  +  82  =  0. 

7.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points  (6,  4,  3)  and  (6,  4,  9),  and  also  from  (—  5,  8,  3)  and  (—  5,  0,  3),  and 
determine  the  nature  of  the  locus.  Ans.   2  =  6,  y  =  4. 

8.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points  (3,  7,  —  4),  (—  5,  7,  —  4),  and  (—  6,  1,  —  4),  and  determine  the 
nature  of  the  locus.  Ans.   x  =  —  1,  y  =  ^. 

148.  Locus  of  one  equation.    Second  fundamental  problem. 

The  locus  of  one  equation  in  three  variables  (one  or  two  may  be 

lacking)  representing  coordinates  in  space  is  the  surface  passing 

through  all  points  whose  coordinates  satisfy  that  equation  and 

through  such  points  only. 

The  coordinates  of  points  on  the  surface  may  be  obtained  as  follows : 

Solve  the  equation  for  one  of  the  vai'iables,  say  z,  assume  pairs  of  values  of 

a;  and  y,  and  compute  the  corresponding  values  of  z. 

A  rough  model  of  the  surface  might  then  be  constructed  by  taking  a  thin  board 

for  the  XF-plane,  sticking  needles  into  it  at  the  assumed  points  (x,  y)  Avhose 

lengths  are  the  computed  values  of  z,  and  stretching  a  sheet  of  rubber  over  their 

extremities. 

The  second  fundamental  problem,  namely,  of  constructing  the 
locus,  is  usually  discarded  in  space  on  account  of  the  mechanical 
difficulties  involved. 

149.  Locus  of  two  equations.  Second  fundamental  problem. 
The  locus  of  two  equations  in  three  variables  representing  coor- 
dinates in  space  is  the  curve  passing  through  all  points  whose 
coordinates  satisfy  both  equations  and  through  such  points  only. 

The  coordinates  of  points  on  the  curve  may  be  obtained  as  follows : 
Solve  the  equations  for  two  of  the  variables,  say  x  and  y,  in  terms  of  the  third, 
z,  assume  values  for  z,  and  compute  the  corresponding  values  of  x  and  y. 


344 


ANALYTIC  GEOMETRY 


150.  Discussion  of  the  equations  of  a  curve.  Third  funda- 
mental problem.  The  discussion  of  curves  in  Elementary  Ana- 
lytic Geometry  is  largely  confined  to  curves  which  lie  entirely  in 
a  plane  which  is  usually  parallel  to  one  of  the  coordinate  planes. 
Such  a  curve  is  defined  as  the  intersection  of  a  given  surface 
with  a  plane  parallel  to  one  of  the  coordinate  planes.  The  method 
of  determining  its  nature  is  illustrated  in 

Ex.  1.  Determine  the  nature  of  the  curve  in  which  the  plane  2  =  4  inter 
sects  the  surface  whose  equation  is  y^  +  z^  =  4  x. 

Solution.    The  equations  of  the  curve  are,  by  definition, 

(1)  2/2  +  z2  =  4  X,     2;  =  4. 

Eliminate  z  by  substituting  from  the  second  equation  in  the  first.    This  gives 

(2)  *?/'^-4x  +  16  =  0,     z  =  4. 
Equations  (2)  are  also  the  equations  of  the  curve.  ' 

For  every  set  of  values  of  {x,  y,  z)  which  satisfy  both  of  ecLuations  (1)  will  evidently 
satisfy  both  of  equations  (2),  and  conversely. 


If  we  take  as  axes  in  the  plane  2  =  4  the  lines  O'X'  and  O'Y'  in  which  the 
plane  cuts  the  ZX-  and  YZ-planes,  then  the  equation  of  the  curve  when 
referred  to  these  axes  is  the  first  of  equations  (2),  namely, 
(3)  y^  -4tx^\Q  =  0. 

For  the  second  of  equations  (2)  is  satisfied  by  all  points  in  the  plane  of  X^,  O',  and  Y\ 
and  the  first  of  equations  (2)  is  satisfied  by  the  points  in  that  plane  lying  on  %e  curve  (3), 
because  the  values  of  the  first  two  coordinates  of  a  point  are  evidently  the  same  when 
referred  to  the  axes  O'X',  O'  Y',  and  O'Z  as  when  referred  to  the  axes  OX,  O  Y,  and  OZ. 

The  locus  of  (3)  is  a  parabola  (Rule,  p.  197)  whose  vertex,  in  the  plane 
z  =  A,  m  the  point  (4,  0)  for  which  p  =  2. 


SURFACES,  CUKVES,  AND  EQUATIONS  345 

The  method  employed  in  Ex.  1  enables  us  to  state  the 

Rule  to  determine  the  nature  of  the  curve  in  which  a  plane  par- 
allel to  one  of  the  coordinate  planes  cuts  a  given  surface. 

First  step.  Eliminate  the  variable  occun^ng  in  the  equation  of 
the  plane  from  the  equations  of  the  plane  and  surface.  The  result 
is  the  equation  of  the  curve  referred  to  the  lines  in  which  the  given 
plane  cuts  the  other  two  coordinate  planes  as  axes. 

Second  step.  Determine  the  nature  of  the  curve  obtained  in  the 
second  step  by  the  methods  of  Plane  Analytic  Geometry. 

PROBLEMS 

1.  Determine  the  nature  of  the  following  curves  and  construct  their  loci. 

(a)  x2  -  4  ?/2  =  8  z,  z  =  8.  (e)  x2  +  4  2/2  +  9  z2  zr  36,  ?/  =  1. 

(b)  x2  +  9?/2  =  9z*  z  =  2.  (f)  a;2  -  4?/2  +  ^2  =  25,  x  =  -  3. 

(c)  x2-4?/2  =  4z,  2/ =  -2.  (g)  x2-2/2-4z2  +  6x  =  0,  x=2. 

(d)  x2  +  ?/2  +  z2  =  25,  X  =  3.  (h)  2/2  +  z2  -  4  X  +  8  =  0,  y  =  4. 

2.  Construct  the  curves  in  which  each  of  the  following  surfaces  intersect 
the  coordinate  planes. 

(a)  x2  +  4 2/2  +  16 z2  =  64.  (d)  x2  +  92/2  =  10 z. 

(b)  x2  +  4  2/2-16z2  =  64.  (e)  x2-9  2/2=10z. 

(c)  x2  -  4  2/2  -  16  z2  =  64.  (f)  x2  +  4  2/2  -  16  z2  =  0. 

3.  Show  that  the  curves  of  intersection  of  each  of  the  surfaces  in  problem 
2  with  a  system  of  planes  parallel  to  one  of  the  coordinate  planes  are  similar 
conies.    In  what  cases  must' this  statement  be  modified  ? 

4.  Determine  the  nature  of  the  intersection  of  the  surface  x2  +  2/2  +  4  z2  =  64 
with  the  plane  z  =  k.  How  does  the  curve  change  as  k  increas^^  irom  0- 
to  4?  from  —  4  to  0?  What  idea  of  the  appearance  of  tiie  surface  is  thus 
obtained  ? 

6.  Determine  the  nature  of  the  intersection  of  the  surface  4x  —  2  2/  =  4 
with  the  plane  y  =  k;  with  the  plane  z  =  k'.  How  does  the  intersection 
change  as  k  or  ¥  changes  ?    What  idea  of  the  form  of  the  surface  is  obtained  ? 

151.  Discussion  of  the  equation  of  a  surface.  Third  funda- 
mental problem. 

Theorem  III.  The  locus  of  an  algebraic  equation  passes  through 
the  origin  if  there  is  no  constant  term  in  the  equation. 

The  proof  is  analogous  to  that  of  Theorem  VI,  p.  73. 


346 


ANALYTIC  GEOMETRY 


Theorem  IV.  If  the  locus  of  an  equation  is  unaffected  by  chang- 
ing the  sign  of  one  variable  throughout  its  equation,  then  the  locus 
is  symmetrical  with  resj^ect  to  the  coordinate  plane  from  which  that 
variable  is  measured. 

If  the  locus  is  unaffected  by  changing  the  signs  of  two  variables 
throughout  its  equation,  it  is  symmetrical  with  respect  to  the  axis 
along  which  the  third  variable  is  measured. 

If  the  locus  is  unaffected  by  changing  the  signs  of  all  three  variables 
throughout  its  equation,  it  is  symmetrical  with  respect  to  the  origin. 

The  proof  is  analogous  to  tliat  of  Theorem  IV,  p.  72. 

Rule  to  find  the  intercepts  of  a  surface  on  the  axes  of  coordinates. 

Set  each  pair  of  variables  equal  to  zero  and  solve  for  real  values 
of  the  third. 

The  curves  in  which  a  surface  intersects  the  coordinate  planes 
are  called  its  traces  on  the  coordinate  planes.  From  the  first 
step  of  the  Rule,  p.  345,  it  is  seen  that 

The  equations  of  the  traces  of  a  surface  are  obtained  by  succes- 
sively setting  x  =  0,  y  =  0,  and  z  =  0  in  the  equation  of  the  surface. 

By  these  means  we  can  determine  some  properties  of  the  surface. 
The  general  appearance  of  a  surface  is  determined  by  considering 
the  curves  in  which  it  is  cut  by  a  system  of  planes  parallel  to  each 
of  the  coordinate  planes  (Eule,  p.  345).     This  also  enables  us  to 

determine  whether  the  sur- 
face is  closed  or  recedes  to 
infinity. 

Ex.  1.  Discuss  the  locus  of 
the  equation  y^  4-  z'^  =  iz. 

Solution.  1.  Tha  surface 
passes  through  the  origin  since 
there  is  no  constant  term  in 
its  equation. 

2.  The  surface  is  sym- 
metrical with  respect  to  the 
XF-plane,  the  ZX-plane,  and 
the  X-axis. 

For.the  locus  of  the  given  equation  is  unaffected  by  changing  the  sign  of  z,  of  y,  or  of 
both  together. 


SURFACES,  CURVES,  AND  EQUATIONS  347 

3.  It  cuts  the  axes  at  the  origin  only. 

4.  Its  traces  are  respectively  the  point-circle  y^-\-z^  =  0  and  the  parabolas 
z2  =  4  cc  and  y^  =  4:X. 

5.  It  intersects  the  plane  x  =  A:  in  the  curve  (Rule,  p.  345) 

2/2  +  2;2  =  4  A;. 

This  curve  is  a  circle  whose  center  is  the  origin,  that  is,  is  on  the  X-axis, 
and  whose  radius  is  2  Vfc  if  A; > 0,  but  there  is  no  locus  if  k<0.  Hence  the 
surface  lies  entirely  to  the  right  of  the  FZ-plane. 

If  k  increases  from  zero  to  infinity,  the  radius  of  the  circle  increases  from 
zero  to  infinity  while  the  plane  x  =  k  recedes  from  the  FZ-plane. 

The  intersection  of  a  plane  z  =  k  or  y=k',  parallel  to  the  XY-  or  ZX-plane, 
is  seen  (Rule,  p.  345)  to  be  a  parabola  whose  equation  is  (compare  Ex.  1,  p.  344) 

y2  =  4x-fe2    or    z^  =  4:X-k'^. 

These  parabolas  are  found  to  have  the  same  value  of  p,  namely,  p  =  2, 
and  their  vertices  recede  from  the  YZ-  or  ZX-plane  as  k  or  k'  increases 
numerically. 


PROBLEMS 

1 .  Discuss  the  loci  of  the 

following  equations. 

(a)  x2  +  22  =  4  a;. 

(f )    X2  +  2/2  -  22  =  0. 

(b)    x2-fy2  +  422  =  16. 

(g)    X2  -  2/2  _  22  ^  9. 

(C)    X2  +  ?/2  -  4  Z2  =  16. 

(h)  x2  +  2/2-2;2  +  2xy  =  0 

(d)  6x  +  4?/+3z  =  12. 

(i)  x  +  y-6z  =  e. 

(e)  Sx  +  2y  +  z  =  12. 

(j)    2/2  +  22  =  25. 

2.  Show  that  the  locus  of  Ax  +  By  +  Cz  +  D  —  0  is  a.  plane  by  considering 
its  traces  on  the  coordinate  planes  and  the  sections  made  by  a  system  of 
planes  parallel  to  one  of  the  coordinate  planes. 

3.  Find  the  equation  of  the  locus  of  a  point  which  is  equally  distant  from 
the  point  (2,  0,  0)  and  the  FZ-plane  and  discuss  the  locus. 

Ans.   2/2  -1-  22  _  4  X  +  4  =  0. 

14.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
point  (0,  0,  3)  is  twice  its  distance  from  the  XF-plane  and  discuss  the  locus. 
Ans.   x2 +  2/2- 3  22-62  +  9  =  0. 
[ 


5.  Finj^  the  equation  of  the  locus  of  a  point  whose  distance  from  the  point 
(0,  4,  0)  is  three  fifths  its  distance  from  the  ZX-plane  and  discuss  the  locus. 
Ans.   25x2 +  16 2/2 +  2522 -200?/ +  400  =  0. 


CHAPTER   XVIII 

THE  PLANE  AND  THE  GENERAL  EQUATION  OF  THE 
FIRST  DEGREE  IN  THREE   VARIABLES 

152.  The  normal  form  of  the  equation  of  the  plane.   Let 

ABC  be  any  plane,  and  let  ON  be  drawn  from  the  origin  per- 
pendicular to  ABC  at  D.  Let  the  positive  direction  on  ON  he 
from  O  toward  N,  that  is,  /ro7^  the  origin  toward  the  plane,  and 


denote  the  directed  length  OD  by  p  and  the  direction  angles  of 
ON  (p.  330)  by  a,  (3,  and  y.  Then  the  position  of  any  j^^ci.ne 
is  determined  by  given  positive  values  of  p,  a,  (3,  and  y. 

Conversely,  a  given  plane  determines  a  single  set  of  positive  values  of  p,  a,  ^, 
and  7  unless  p  =  0.  lip=  0,  the  positive  direction  on  ON  becomes  meaningless. 
Ifp  —  0,  we  shall  suppose  that  ON  is  directed  upward,  and  hence  cos  7  >  0  since 

It 
7  <  —  •     If  the  plane  passes  through  OZ,  then  ON  lies  in  the  XF-plane  and 

2  It 

cos  7  =  0;  in  this  case  we  shall  suppose  ON  so  directed  that  /3  <  17  and  hence 

cos  /3  >  0.  Finally,  if  the  plane  coincides  with  the  TZ-plane,  the  positive  direction 
on  ON  shall  be  that  on  OX. 

348 


THE  PLANE  349 

Theorem  I.    Normal  form.    TJie  equation  of  a  plane  is 

(I)  X  cos  a  +  2/  cos  ^  +  5!  cos  y  —  jp  =  O, 

where  p  is  the  perpendicular  distance  from  the  origin  to  the  plane, 
and  a,  ^,  and  y  are  the  direction  cosines  of  that  perpendicular. 

Proof    Let  P(x,  y,  z)  be  any  point  on  the  given  plane  ABC. 
Project  OEFP  and  OP  on  the  line  ON.     By  Theorem  II,  p.  328, 

proj.  of  OE  +  proj.  of  EF  +  proj.  of  FP  =  proj.  of  OP. 

Then  by  Theorem  I,  p.  328,  and  by  the  definition,  p.  29, 

X  cos  a  +  y  cos  ^  +  z  cos  y  =  p. 

Transposing,  we  obtain  (I).  q.e.d. 

Corollary.    The  equation  of  any  plane  is  of  the  first  degree  in 

X,  y,  and  z. 

153.  The  general  equation  of  the  first  degree,  Aoc  +  By  +  Cz 
+  1>  =  0. 

Theorem  II.  (Converse  of  the  Corollary.)  The  locus  of  the  gen- 
eral equation  of  the  first  degree  in  x,  y,  and  z, 

(II)  Aic-\-  By  -\-Cz  +  D  =  0, 

is  a  plane. 

Proof  We  shall  prove  the  theorem  by  showing  that  (II)  may' 
be  reduced  to  the  form  (I)  by  multiplying  by  a  proper  constant. 
To  determine  this  constant,  multiply  (II)  by  k,  which  gives 

(1)  kAx  +  kBy  +  kCz  +  kD  =  0. 
Equating  corresponding  coefficients  of  (1)  and  (I),  we  get 

(2)  kA  =  cos  a,     kB  =  cos  (3,     kC  =  cos  y,     kD  =  —p. 
Squaring  the  first  three  of  equations  (2)  and  adding, 

k^(A^  -\-B^  -{-  C^)  =  cos^a  +  cos^^  +  cos^y  =  1. 

(by  (III),  p.  330) 

(3)  .\  k= ^- 


350  ANALYTIC  QEOMETRY 

From  the  last  of  equations  (2)  we  see  that  the  sign  of  the 

radical  must  be  opposite  to  that  of  D  in  order  that  p  shall  be 

positive. 

If  Z)  =  0,  then  p  =  0;  and  from  the  third  of  equations  (2)  the  sign  of  the  radical 
must  be  the  same  as  that  of  C,  since  when  p  =  0  cos  7  >  0.  If  D  =  0  and  C  =  0, 
then  p—0  and  cos  7=0 ;  and  from  the  second  of  equations  (2)  the  sign  of  the  radical 
must  be  the  same  as  that  of  B,  since  when  p  —  0  and  cos  7  =  0  cos  /3  >  0. 

Substituting  from  (3)  in  (2),  we  get 


COS  «  = ,  )     COS  B  = i 

C  -D 

COS  y 


We  have  thus  determined  values  of  a,  /?,  y,  and  p  such  that  (I) 
and  (II)  have  the  same  locus.  Hence  the  locus  of  (II)  is  a 
plane.  q.e.d. 

Corollary  I.  The  direction  cosines  of  a  normal  to  the  plane  (II) 
are  respectively  A,  B,  and  C  each  divided  hy  ±  Vvl^  -\-  B'^  -\-  C^. 
The  sign  of  the  radical  is  opposite  to  that  of  D,  the  same  as  that  of 
C  if  D  =  0,  the  same  as  that  of  B  if  C  =  D  =  0,  or  the  same  as 
that  of  A  ifB  =  C  =  D  =  0. 

Corollary  II.  To  reduce  the  equation  of  a  plane  to  the  normal 
form  divide  its  equation  hy  zt  V^^  -\-  B'^  -\-  C^,  choosing  the  sign  of 
the  radical  as  in  Corollary  I. 

Corollary  III.     Two  planes  whose  equations  are 

Ax  +  By  +  Cz  +  D  =  0,     A'x  +  B'y  -i-  C'z  -[-  D'  =  0 
are  parallel  when  and  only  when  the  coefficients  of  x,  y,  and  z  are 

proportional,  that  is, 

A__B__C^ 

A'~B'~C'' 

For  from  Corollary  I  the  direction  cosines  of  a  normal  to  (II)  are  proportional 
to  A,  B,  and  C,  and  two  planes  are  evidently  parallel  when  and  only  when  their 
normals  are  parallel  (Corollary,  p.  335) . 

Corollary  IV.    Two  planes  are  perpendicular  when  and  only  when 

AA'  ^BB^  ^-  CC  =  0. 

This  follows  from  Corollary  I  by  the  Corollary  on  p.  335,  since  two  planes  are 
perpendicular  when  and  only  when  their  normals  are  perpendicular. 


THE  PLANE  351 

Corollary  V.    A  plane  whose  equation  has  the  form 

Ax  -{-  By  -\-  D  =  0  is  perpendicular  to  the  XY-plane; 
By  -\-  Cz  -{-  D  z=  0  is  perpendicular  to  the  YZ -plane; 
Ax-\-  Cz  -\-  D  =  (i  is  perpendicular  to  the  ZX-plane. 

That  is,  4f  one  variable  is  lacking,  the  plane  is  perpendicular  to 

the  coordinate  plane  corresponding  to  the  two  variables  which  occur 

in  the  equation. 

For  these  planes  are  respectively  perpendicular  to' the  planes  2;  =  0,  x  =  0,  and 
y  =  0  by  Corollary  IV. 

Corollary  VI.    A  plane  whose  equation  has  the  form 

Ax  -^  D  =  0  is  perpendicular  to  the  axis  of  x; 
By  +  D  =  0  is  perpendicular  to  the  axis  of  y ; 
Cz  -{-  D  =  0  is  perpendicular  to  the  axis  of  z. 

That  is,  if  two  variables  are  lacking,  the  plane  is  perpendicular  to 
the  axis  corresponding  to  the  variable  which  occurs  in  the  equation. 

For  by  Corollary  I  two  of  the  direction  cosines  of  the  normal  to  the  plane  are 
zero  and  hence  the  normal  is  parallel  to  one  of  the  axes  and  the  plane  is  therefore 
perpendicular  to  that  axis. 

PROBLEMS 

1.  Eind  the  intercepts  on  the  axes  and  the  traces  on  the  coordinate  planes 
of  each  of  the  following  planes  and  construct  the  figures. 

(a)  2ic  +  3?/  +  4z  -  24  =  0.  (e)  5a;  -  7y  -  35  =  0. 

Xb)  Tx-Sy +  2-21=0.  (f)  4X  +  32; +36  =  0. 

■^^  (c)  9x-7y-9z  +  6S  =  0.  (g)  5?/ -  82  -  40  =  0. 

(d)  6x  +  4y -2+ 12  =  0.  (h)  3x  +  52  +  45  =  0. 

2.  Find  the  equations  of  the  planes  and  construct  them  by  drawing  their 
traces,  for  which 

Ans.   V2  X  +  2/  +  z  -  12  =  0. 

Ans.   x  +  V2y-2;  +  16  =  0. 
Ans.    6x-2y  +  Sz-2Sz=0. 
Ans.   2x  +  y  +  22  +  6  =  0. 


,p  =  e 

(b)a  =  ?^,^  =  5^,.= 

It 
=  3'^ 

cos  a      cos/3      C0S7 
^'^      6     -  -2          3     ' 

1>  =  4, 

,,,  cos  or      cosjS      cos  7 
(^)    _2   =   -l  =   -2' 

p  =  2, 

352  ANALYTIC  GEOMETRY 

3.  Find  the  equation  of  the  plane  such  that  the  foot  of  the  perpendicular 
from  the  origin  to  the  plane  is  the  point 

(a)  (-3,- 2,  a).  Ans.    3x  -  2  ?/ -  62  +  49  =  0. 

(b)  (4,  3,  -  12).  Ans.    4  x  +  3  ?/  -  12  z  -  169  =  0. 

(c)  (2,  2,  -  1).  Ans.    2x  +  2y  -z-9  =  0. 

4.  Reduce  the  following  equations  to  the  normal  form  and  find  a-,  j8,  7, 
and  p. 

(a)  6x-Sy  +  2z-7  =  0.       Ans.    cos-if,  cos-i(- f),  cos-if,  1. 

(b)  x-V2y  +  z  +  8  =  0.  Ans.    ?^,  j,  ^,  4. 

o       4       o 

(c)  2x-2y -z +  12  =  0.         Ans.    cos-i(- |),  cos-if,  cos-ii,  4. 

(d)  y-z +  10  =  0.  Ans.   -,  —  ,  -,  5V2. 

(e)  Sx-{-2y  -6z  =  0.  Ans.    cos-i(- f),  cos-i(-f),  cos-if,  0. 

5 .  Find  the  distance  from  the  origin  to  the  plane  12x  —  4?/  +  3z  —  39  =  0. 

Ans.    3. 

6.  Find  the  distance  between  the  parallel  planes  6x  +  2y  —  Sz  —  6Pi  =  0 
and6x  +  22/-32;  +  49  =  0.  Ans.  16. 

7.  What  may  be  said  of  the  position  of  the  plane  (I)  if 

(a)  cos  a  =  0?  (c)   cos  7  =  0  ?  (e)  cos  j3  =  cos  7  =  0  ? 

(b)  cos  /3  =  0  ?  (d)  cos  «  =  cos/3  =  0  ?  (f )  cos  7  =  cos  or  =  0  ? 

8.  What  are  the  equations  of  the  traces  on  the  coordinate  planes  of  the 
plane  Ax  +  By  +  Cz  -{■  D  =  0? 

9.  Show  that  the  following  pairs  of  planes  are  either  parallel  or  perpen- 
dicular. 

J2x-hBy-ez  +  S  =  0,  fex-Sy  +  2z-1  =0, 

W  \6x+l^y-lSz-6  =  0.  ^^'  \sx  +  2y-6z-\-28  =  0. 

3x-5?y-4z  +  7  =  0,  ,^,^    (Ux  -  7  y  -  21  z  -  50  =  0, 

3z  +  12  =  0. 


fSx-6y-4z  +  7  =  0,  ri4x-7; 

^  ^  \6x  +  2y  +  2z-7  =  0.  ^  ^  \  2x-y 


10.  For  what  values  of  a,  )3,  7,  and  p  will  the  locus  of  (I)  be  parallel 
to  the  XF-plane  ?  the  YZ-plane  ?  the  ZX-plane  ?  coincide  with  each  of 
these  planes  ? 

11.  For  what  values  of  a,  ^,  7,  and  p  will  the  locus  of  (I)  pass  through 
the  X-axis  ?    the  F-axis  ?    the  Z-axis  ? 

12.  Show  that  the  coordinates  of  the  point  of  intersection  of  three  planes 
may  be  found  by  solving  their  equations  simultaneously  for  x,  y,  and  z. 


THE  PLANE  353 

13.  Find  the  coordinates  of  the  point  of  intersection  of  the  planes 
x-{-2y  -{■  z  =  0,  X  -2y  -8  =  0,  and  x  +  y-\-z-S  =  0. 

Ans.    (2,  -  3,  4). 

14.  Show  that  the  plane  x  +  2y  —  2z  —  9  =  0  passes  through  the  point 
of  intersection  of  the  planes  x  +  y-^z  — 1  =  0,  x—y  —  z  —  l  =  0,  and 
2xi-Sy  -8  =  0. 

15.  Show  that  the  four  planes  x  +  y  +  2z  —  2  =  0,  x  +  y  —  2z  -\- 2  =  0, 
X  —  y  +  8  =  0,  and  3x  —  y-2z  +  18  =  0  pass  through  the  same  point. 

16.  Show  that  the  planes  2x  —  y-\-z  +  S  =  0,  x  —  ?/  +  4z  =  0,  Sx  +  y 
-2z  +  8  =  0,^x-2y  +  2z-5  =  0,9x  +  Sy-6z-T=0,a,nd7x-7y 
-\-  28z  —  6  =  0  bound  a  parallelopiped. 

17.  Show  that  the  planes  Qx  —  Sy  +  2z  =  i,  Sx  +  2y  -6z  =10,  2x  +  6y 
+  32  =  9,  Sx  +  2y  -6z  =  0,  12x  +  36y  +  18z-ll  =  0,  and  I2x -6y 
+  42  —  17  =0  bound  a  rectangular  parallelopiped. 

18.  Show  that  the  planes  x+2y  —  z  =  0,  y+1  z—2=0,  Xj2)f  —  z—A=0, 
2x  +  y  —  8  =  0,  and  3x  +  3y  —  z  —  8  =  0  bound  a  quadranaj^lar  pyramid. 

19.  Derive  the  conditions  for  parallelism  of  two  plane^'from  the  fact  that 
two  planes  are  parallel  if  their  traces  are  parallel  lines.  / 

/ 

154.  Planes  determined  by  three  conditions.   If  three  of  the 
coefficients  of 
(1)        .  Ax+By  -{-Cz^D  =  0 

are  known  in  terms  of  the  fourth,  then  the  plane  is  completely 
determined,  for  if  their  values  be  substituted  in  (1),  the  equation 
may  be  divided  by  the  fourth  coefficient.  Three  conditions  which 
the  plane  satisfies  will  lead  to  three  equations  in  the  coefficients 
which  may  be  solved  for  three  of  the  coefficients  in  terms  of  the 
fourth.  Hence  a  plane  is,  in  general,  determined  by  three  con- 
ditions. Its  equation  may  be  obtained  by  a  Rule  analogous  to 
that  on  p.  93,  using  equation  (1)  in  the  first  step. 

Thus  to  find  the  equation  of  a  plane  passing  through  three  points  we  proceed 
as  in  Ex.  1,  p.  93,  using  equation  (1)  in  the  first  step.  In  the  second  step  three 
equations  involving  A,  B,  C,  and  7)  are  obtained,  which  may  be  solved  for  three 
of  these  coefl&cients  in  terms  of  the  fourth. 


354 


ANALYTIC   GEOMETRY 


Ex.  1.    Find  the  equation  of  the  plane  which  passes  through  the  point 
Pi  (2,  -7,  I)  and  is  parallel  to  the  plane  21  ic  -  12  ?/  +  28  z  -  84  =  0. 
Solution.    Let  the  equation  of  the  required  plane  be 

(2)  Ax  +  By-{-Cz  +  D  =  0. 

Since  Pi  lies  on  (2), 

(3)  2A-7B  +  IC  +  D  =  0, 

and  since  (2)  is  parallel  to  the  given  plane  (Corollary  III,  p.  350), 
A  _  Ji^  _  C^ 
21  ~  -12"  28* 


(4) 


Solving  (3)  and  (4)  for  J.,  J5,  and  D  in  terms  of  C,  we  get 

A  =  ^C,     B  =  -^C,     D=-6C.- 
Substituting  in  (2),  we  obtain 

lCx-^Cy  +  Cz-6C  =  0. 
Clearing  of  fractions  and  dividing  by  C, 

21 X  -  12  2/  +  28  z  -  168  =  0. 


PROBLEMS 

1.  Find  the   equation   of  the  plane   which  passes  through   the  points 
(2,  3,  0),  (-2,  -3,  4),  and  (0,  6,  0).  Ans.    Sx  +  2y -{- 6z -12  =  0. 

\y   2.  Find  the   eqiiation   of  the  plane   which  passes  through  the  points 
"  1,  i,  -1),  (-2,  -2,  2),  and  (1,  -1,  2).  Ans.    x-Sy  -2z  =  0. 

3.  Find  the  equation   of  the  plane   which    passes   through    the    point 
(3,  —  3,  2)  and  is  parallel  to  the  plane  Sx  —  y  +  z  —  6  =  0. 

Ans.   3x  —  y-\-z  —  14:  =  0. 


\) 


THE  PLANE  355 

4.  Find  the  equation  of  the  plane  which  passes  through  the  points 
(0,  3,  0)  and  (4,  0,  0)  and  is  perpendicular  to  the  plane  4x*—  6y  —  z  =  12. 

Ans.   Sx  +  iy  -12z-l2  =  0. 

5.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(0,  0,  4)  and  is  perpendicular  to  each  of  the  planes  2x  —  Sy  =  b  and 
x-iz  =  3.  Ans.    12a: +  82/ +  32 -12  =  0. 

6.  Find  the  equation  of  the  plane  whose  intercepts  on  the  axes  are 
3,  5,  and  4.  Ans.    20x  +  12?/ +  15  2  -  60  =  0. 

\/t.  Find   the   equation  of  the  plane  which   passes  through   the  point 
(2,  —  1,  6)  and  is  parallel  to  the  plane  a>—  2^  —  32  +  4  =  0. 

Ans.    x-2y-3z  +  14:  =  0. 

8,  Find  the  equation  of  the  plane  which  passes  through  the  points 
(2,  —1,  6)  and  (1,  -2,  4)  and  is  perpendicular  to  the  plane  x  — 2?/  —  2  2  +  9=0. 

Ans.   2x  +  4y-32;  +  18  =  0. 

9.  Find  the  equation  of  the  plane  w^ose  intercepts  are  —1,  —1,  and  4. 

Ans.    4x  +  42/-2  +  4=0. 

10.  Find  the  equation  of  the  plane  which  passes  through  the  point 
(4,  —2,  0)  and  is  perpendicular  to  the  planes  x+y—z=0  and  2x— 4y  +  z=5. 

Ans.   x-\-y  +  2z-2  =  0. 

11.  Show  that  the  four  points  (2,  -3,4),  (1,0,2),  (2,  -1,2),  and 
(1,  —  1,  3)  lie  in  a  plane. 

12.  Show  that  the  four  points  (1,  0,  -1),  (3,  4,  -3),  (8,  -2,  6),  and 
(2,  2,  —  2)  lie  in  a  plane. 

13.  Find  the  equation  of  the  plane  which  is  perpendicnlai"  to  the  line 
joining  (3,  4,  —  1)  to  (5,  2,  7)  at  its  middle  point.      4iP^  '*  \     . 

Ans.   x  ^'t^  2  -^  13  =  0. 
^'^  i        \ 

14.  Find  the  equations  of  the  faces  of  the  tetraedraiBf^hose  v^tices^rfe' 
the  points  (0,  3,  1),  (2,  -  7,  1),  (0,  5,  -  4),  and  (2,  0,  l).'' 

Ans.    25x  +  52/ +  22  =  17,  5x- 22  =  8,  2  =  1,  15x  +  lOy  +  42  =  34. 

15.  The  equations  of  three  faces  of  a  parallelepiped  are  x  — 4?/  =  3, 
2x-2/  +  2  =  3,  and  3x  +  y-22  =  0,  and  one  vertex  is  the  point  (3,  7,  -  2). 
What  are  the  equations  of  the  other  three  faces  ? 

'  Ans.   x-42/  +  25  =  0,  2x-?/  +  2  +  3  =  0,  3x  +  ?/-22  =  20. 


16.  Find  the  equation  of  the  plane  whose  intercepts  are  a,  6,  c 

y 

b 


Ans.    -  +  -  +  - 


V- 


356  ANALYTIC  GEOMETRY 

17.  What  are  the  equations  of  the  traces  of  the  plane  in  problem  16? 
How  might  these  equations  have  been  anticipated  from  Plane  Analytic 
Geometry  ? 

18.  Find  the  equation  of  the  plane  which  passes  through  the  point 
-Pi  {^ii  2/i5  Zi)  and  is  parallel  to  the  plane  AiX  +  Biy  +  dz  +  Di  =  0. 

Ans.   Ai (X  -  Xi)  +  Bx{y  -  yi)  +  Ci (2  -  Zi)  =  0. 

19.  Find  the  equation  of  the  plane  which  passes  through  the  origin  and 
-Pi  i^u  2/1?  ^1)  a-nd  is  perpendicular  to  the  plane  AiX  +  Biy  +  CiZ  +  Di  =  0. 

Ans.   {Bizi  -  Ciyi)x  +  (CiXi  -  AiZi) y  +  {AiVi  -  BiXi) z  =  0. 

155.  The  equation  of  a  plane  in  terms  of  its  intercepts. 

Theorem  III.  If  a,  b,  and  c  are  the  intercepts  of  a  plane  on  the 
X-j  Y-,  and  Z-axes  respectively,  then  the  equation  of  the  plane  is 

(III)  ^  +  ^  +  ^  =  1.  ? 

Proof  By  Theorem  II  the  equation  of  any  plane  has  the  form 
(1)  Ax-\-Bij  -\-Cz^D  =  0. 

By  the  Rule,  p.  346,  we  get 

D 

^  D 

whence  A= 5   B=  —  —i    C= 

a  0  G 

Substituting  in  (1),  dividing  by  —  D,  and  transposing,  we 
obtain  (III)-  q.e.d. 

Equation  (III)  should  be  compared  with  (VI),  p.  96. 

156.  The  distance  from  a  plane  to  a  point.  The  positive  direc- 
tion on  any  line  perpendicular  to  a  plane  is  assumed  to  agree  with 
that  on  the  line  drawn  through  the  origin  perpendicular  to  the 
plane  (p.  348).  Hence  the  distance  from  a  plane  to  the  point  Pi 
is  positive  or  negative  according  as  Pi  and  the  origin  are  on  oppo- 
site sides  of  the  plane  or  not. 

If  the  plane  passes  through  the  origin,  the  sign  of  the  distance  from  the  plane 
to  Pi  must  be  determined  by  the  conventions  for  the  special  cases  on  p.  348. 


D 

—  F 

a 

"-T 

THE  PLANE  357 

Theorem  IV.    The  distance  d  from  the  plane 

X  cos  cc  -\-  y  cos  y8  +  ^  cos  y  —  p  =  0 
to  the  point  P^  (x-i,  yi,  z-^)  is  given  by  ^■ 

(IV)  ci  =  Xi  cos  a  +  2/1  cos  p  +  z^cosy  —  p. 

Proof.    Projecting  OP^  on  ON,  we  evidently  get  jo  +  d. 
Projecting  OE,  EF,  and  FP^  on  ON,  we  get  respectively  (Theo- 
rem I,  p.  328)  Xi  cos  a,  2/1  cos  (3,  and  Zi  cos  y. 


Then  by  Theorem  II,  p.  328, 

p  -\-  d  z=  Xi  cos  a  +  2/1  cos  y8  H-  ^1  cos  y. 

.-.  d  =  XiGOSa  -\-  2/1  cos  ft  -\-  Zi  cos  y  —p.  Q.E.D. 

Prom  Theorem  lY  we  have  at  once  the 

Rule  to  find  the  distance  from  a  given  plane  to  a  given  point. 

First  step.  Reduce  the  equation  of  the  plane  to  the  normal  form 
{Corollary  II,  p.  S50). 

Second  step.  Substitiite  the  coordinates  of  the  given  point  in  the 
left-hand  side  of  the  equation.     The  result  is  the  required  distance. 

157.  The  angle  between  two  planes.  The  plane  angle  of  one 
pair  of  diedral  angles  formed  by  two  intersecting  planes  is  evi- 
dently equal  to  the  angle  between  the  positive  directions  of  the 
normals  to  the  planes.     That  angle  is  called  the  angle  between  the 


358  ANALYTIC  GEOMETRY 

Theorem  V.    The  angle  6  hetiveen  two  planes 

is  given  by 


^7ie  si(7W5  o/  ^Ae  radicals  being  chosen  as  in  Corollary  I,  p.  350. 

Proof.  By  definition  the  angle  between  the  planes  is  the  angle 
between  their  normals. 

By  (4),  p.  350,  the  direction  cosines  of  the  normals  to  the  planes 
are 

cos  oTi  = ,       "''  ?      cos  cr. 

cos  Pi  = .  ^  ?      cos  ^2  = 


^ii 

±^A, 

+  c\' 

±Va{ 

'  +  B,' 

c. 

+  Ci^ 

±vi; 

'  +  B, 

'^ci 

±Va, 

'+B^' 

'^c^ 

C, 

cos  yi  = ,  ?       COS  72  .  

By  (V)^  P-  334,  we  have 

COS  ^  =  COS  «i  COS  CTs  +  COS  ^1  COS  ^2  +  COS  yi  COS  ya. 

Substituting  the  values  of  the  direction  cosines  of  the  normals, 
we  obtain  (V).  q.e.d. 

PROBLEMS 

1.  Find  the  distance  from  the  plane 

(a)  6x  -  3  ?/  +  2  2:  -  10  =  0  to  the  point  (4,  2,  10).  Ans.  4. 

(b)  X  +  2  2/  -  2  z  -  12  =  0  to  the  point  (1,  -  2,  3).  Ans.  -  7. 

(c)  4 X  4-  3 ?/  +  12  2;  +  6  =  0  to  the  point  (9,  -1,0).  Am.  -  3. 

(d)  2x  -  5?/  +  3  2;  -  4  =  0  to  the  point  (-2,  1,  7).  Ans.  y%  V38. 

2.  Do  the  origin  and  the  point  (3,  5,  —  2)  lie  on  the  same  side  of  the 
plane  7x  —  2/  —  32  +  6  =  0?  An&.   Yes. 

3.  Find  the  distance  from  the  plane  ^x  +  %  +  Cz  +  D  =  0  to  the  point 
-Pi(a^i,  2/1,  ^i)-  ^^^    Axx  +  Byx  +  Czx  +  D 


THE  PLANE  '  359 

4.  Find  the  locus  of  points  which  are  equally  distant  from  the  planes 
2x-y  -2z-3  =  0and6x-Sy  +  2z-\-4i  =  0. 

Ans,   32x-16y-82-9  =  0. 

5.  Find  the  length  of  the  altitude  of  the  tetraedron  whose  vertices  are 
(0,  3,  1),  (2,  -  7,  1)  (0,  5,  -  4),  and  (2,  0,  1)  which  is  drawn  from  the  first 
vertex.  Ans.   ^^V29. 

6.  Find  the  volume  of  the  tetraedron  whose  vertices  are  (3,  4,  0), 
(4,  -  1,  0),  (1,  2,  0),  and  (6,  -  1,  4).  Ans.   8. 

7.  Find  the  angles  between  the  following  pairs  of  planes. 

(a)  2x  +  y  -2z-9  =  0,x-2y  -\-2z  =  0.  Ans.    cos-i(-f). 

(b)  x-\-y-4.z  =  0,Sy-3z  +  7  =  0.  Ans.   cos-  1 1. 

(c)  ^x  +  2y  -{-4z-7  =  0,  3x~4y  =  0.  Ans.    cos-i(-^2^). 

{d)2x-y-^z  =  7,x-\-y  +  2z  =  ll.     .  Ans.   -. 

o 

8.  Show  that  the  angle  given  by  (V)  is  that  angle  formed  by  the  planes 
which  does  not  contain  the  origin. 

9.  Find  the  vertex  and  the  diedral  angles  of  that  triedral  angle  formed  by 
the  planes  x  +  y  +  z  =  2^x  —  y  —  2z=^^  and  2x-\-y  —  z  =  2\n  which  the 

origialies.  ^^    ^^_  _^^  2),  cos-.iVi,  ^,  cos->(-l%^). 

o  o  \      3        / 

^    10.  Find  the  equation  of  the  plane  which  passes  through  the  points 

(0,  —1,  0)  and  (0,  0,  —  1)  and  which  makes  an  angle  of  —  with  the  plane 

y  +  z  =  l.  r-       ^ 

Ans.    ±V6x  +  y  +  2;  +  l  =  0. 

11.  Find  the  locus  of  a  point  which  is  3  times  as  far  from  the  plane 
Zx  —  Qy  —  2z  =  Q  as  from  the  plane  2x  —  y  +  2z  =  ^. 

Ans.   17x-13y  +  12z-63  =  0. 

158.  Systems  of  planes.  The  equation  of  a  plane  which  satis- 
fies two  conditions  will,  in  general,  contain  an  arbitrary  constant, 
for  it  takes  three  conditions  to  determine  a  plane.  Such  an  equa- 
tion therefore  represents  a  system  of  planes. 

Systems  of  planes  are  used  to  find  the  equation  of  a  plane 
satisfying  three  conditions  in  the  same  manner  that  systems  of 
lines  are  used  to  find  the  equation  of  a  line  satisfying  two  condi- 
tions (Eule,  p.  114). 


360  ANALYTIC  GEOMETRY 

Theorem  VI.    The  system  of  planes  parallel  to  a  given  plane 

Ax+Bij  -^Cz  +  D^^) 
is  represented  by 

(VI)  Ax  +  By  -\-Cz-]-h  =  0, 
where  k  is  an  arbitrary  constant. 

Hint.  Show  that  all  of  the  planes  (VI)  are  parallel  to  the  given  plane  by  Corollary  III, 
p.  350,  and  that  every  plane  parallel  to  the  given  plane  is  represented  by  (VI),  by  finding 
a  value  of  k  for  which  (VI)  passes  through  a  given  point  P^. 

Theorem  VII.  The  system  of  planes  passing  through  the  line  of 
intersection  of  two  given  planes 

A^x  +  B^y  +  C^z  +  Z)i  =  0,     A^x  +  B^  +  Cg^  +  A  =  0 

is  represented  by 

(VII)  A^oc  +  Biu  +  C^z  +  I>i  +  A;  {A^x  +  B^y  +  C^z  +  D^)  =  O, 
where  k  is  an  arbitrary  constant. 

Hint.  Show  that  (VII)  passes  through  any  point  on  the  intersection  of  the  given  planes, 
and  find  a  value  of  k  for  which  (VII)  passes  through  any  point  not  on  the  intersection. 

Theorem  VIII.  If  the  equations  of  the  planes  in  Theorem  VII  are 
in  normal  form,  then  —  k  is  the  ratio  of  the  distances  from  those 
planes  to  any  point  in  (^VII). 

Hint.  Let  P-^  (a;,,  y^,  z^)  be  any  point  on  the  plane 

X  cos  a^  4-  2/  cos  Pi  +  s  cos  y^-  Pi  +  k (x  cos  a^  +  y  cos  ^2  +  ^  cos  yg  -  ^2)=  0. 
Then  x^  cos  a^  +  y^  cos  ^^  +  z^  cos  yi  -  i?i  +  k  (x-^  cos  a^  +  2/1  cos  p^  +  ^1  cos  yj  —  Pz)  =  ^^ 
Solve  for  A;  and  interpret  the  result  by  Theorem  IV,  p.  357. 

Corollary.  The  equations  of  the  planes  bisecting  the  angles  for  7ned 
by  two  giveii  planes  are  found  by  reducing  their  equations  to  the 
normal  form  and  adding  and  subtracting  them. 

The  plane  (VII)  will  lie  in  the  external  or  internal  angles 
(p.  121)  formed  by  the  given  planes  according  as  k  is  positive  or 
negative. 

The  equation  of  a  system  of  planes  which  satisfy  a  single  con- 
dition must  contain  two  arbitrary  constants.  One  of  the  most 
important  systems  of  this  sort  is  given  in 


THE  PLANE  361 

Theorem  IX.  The  system  of  planes  passing  through  a  given  point 
Pi  (xi,  )/i,  Zi)  is  represented  by 

(IX)  A(oo-aOi)  +  B(y-yi)-\-C{z-Zi)=0. 

Proof.    Equation  (IX)  is  the  equation  of  a  plane  which'passes 
through  Pi,  for  the  coordinates  of  Pi  satisfy  (IX). 
If  any  plane  whose  equation  is 

Ax  -\-  By  -{-  Cz  -\-  D  =  0 

passes  through  Pj,  then 

Axi  +  %i  +  C^,  +  p  =  0. 

Subtracting,  we  get  (IX).  Hence  (IX)  represents  all  planes 
passing  through  Pi.  q.e.d. 

Equation  (IX)  contains  two  arbitrary  constants,  namely,  the  ratio  of  any  two 
coefficients  to  the  third. 

PROBLEMS 

1.  Determine  the  vakie  of  k  such  that  the  plane  x-\-ky  —  2z  —  ^  =  0  shall 

(a)  pass  through  the  point  (5,  —  4,  —  6).  Ans.  2. 

(b)  be  parallel  to  the  plane  6a;  —  2?/  —  12 2  =  7,  Ans.  —  \. 

(c)  be  perpendicular  to  the  plane  2x  —  42/4-2  =  3.  Ans.  0. 

(d)  be  3  units  from  the  origin.  Ans.  ±  2. 

(e)  make  an  angle  of  —  with  the  plane  2x  —  2y -\-  z  —  0.     Ans.   —  f  V35. 

o 

2.  Find  the  equation  of  the  plane  which  passes  through  the  point  (3,  2,  —  1) 
and  is  parallel  to  the  plane  7  x  —  y  +  z  =  14. 

Ans.    7x  —  y-^z  —  IS  =  0. 

3.  Find  the  equation  of  the  plane  which  passes  through  the  intersection 
of  the  planes  2x  +  y  —  i  =  0  and  2/  +  2 z  =  0  and  which  (a)  passes  through 
the  point  (2,  —  1,  1);  (b)  is  perpendicular  to  the  plane  Sx  -\- 2y  —  Sz  =  6. 

Ans.    (a)  cc +  y  +  2 -2  =  0;    (b)  2x -f  3?/ +  42  -  4  =  0. 

4.  Find  the  equations  of  the  planes  which  bisect  the  angles  formed  by  the 
planes  2x  —  y  +  2z  =  0  and  x-\-2y  —  2z  =  6.  ^ 

Ans.    3x  +  2/-6  =  0,  x-3y +  4z  +  6  =  0. 

5.  Find  the  equations  of  the  planes  passing  through  the  intersection  of  the 
planes  2x  +  y  —  z  =  4:  and  x  —  y  -\-2z  =  0  which  are  perpendicular  to  the 
coordinate  planes.  Ans.    5ic  +  y  =  8,  3x  +  z  =  4,  3?/  —  52=4, 


362  ANALYTIC  GEOMETKY 

6.  Find  the  equations  of  the  planes  which  bisect  the  angles  formed  by  the 
planes  6x  —  2y  —  3z  =  0  and  ix-\-Sy  —  ISz  =  10,  and  verify  by  means  of  (V). 

7.  Find  the  equation  of  the  plane  passing  through  the  intersection  of  the 
planes  AiX  +  Biy  +  C12;  +  Di  =  0  and  A^x  +  ^22/  +  C22;  +  A  =  0  which 
passes  through  the  origin. 

Arts.    (JliDg  -  A2B1)  X  +  {B1B2  -  A^Bi)  y  +  (CiA  -  C2B1)  z  =  0. 

8.  Find  the  equations  of  the  planes  which  bisect  the  angles  formed  by  the 
planes  ^ix  +  Biy  +  Ciz  +  Di  =  0  and  A^x  +  Biy  +  C^z  +  D2  =  0. 

Aix  +  Biy  +  Ciz  +  Di  _      A^x  +  Biy  +  C^z  +  B^ 

-a.7lS. .  = —  ±  — • 

V^i2  +  5^2  +  Ci^  VvlaS  +  ^a^  +  C22 

9.  Find  the  equations  of  the  planes  passing  through  the  intersection  of 
the  planes  AiX  +  B^y  4-  C\Z  +  Di  =  0  and  A^x  +  B^y  +  C2Z  +  I>2  =  0  which 
are  perpendicular  to  the  coordinate  planes. 

Ans.  (^iBg  -  AiBx)  y  -  (Ci^2  -  C^Ai)  z  +  A^B^  -  A^Bi  =  0, 
{A^Bi  -  A2B{)x  -  {B1C2  -  B2C1)  z  -  {B1B2  -  B2B1)  =  0, 
{C1A2  -  C2Ai)x  -  {B1C2  -  B2Ci)y  +  C1B2  -  C2B1  =  0. 

10.  Find  the  equation  of  the  plane  which  passes  through  Pi  (xi,  yi,  Zi) 
and  is  perpendicular  to  the  planes 

Aix  +  Biy  4-  Ciz  +  X)i  =  0  and  A2X  +  -B22/  +  C^z  +  D2  =  0. 
Ans.  {BiG2-B2Ci){x-Xi)+{CiA2-C2Ai){y-yi)  +  {AiB2-A2Bi){z-Zi)=0. 


CHAPTER   XIX 

THE  STRAIGHT  LINE  IN  SPACE 

159.  General  equations  of  the  straight  line.  A  straight  line 
may  be  regarded  as  the  •  intersection  of  any  two  planes  which 
pass  through  it.  The  equations  of  the  planes  regarded  as  simul- 
taneous are  the  equations  of  the  line  of  intersection,  and  hence 
(Corollary,  p.  349) 

Theorem  I.  The  equations  of  the  straight  line  are  of  the  first 
degree  in  x,  y,  and  z. 

Conversely,  the  locus  of  two  equations  of  the  first  degree  is 
a  straight  line  unless  the  planes  which  are  the  loci  of  the  separate 
equations  are  parallel.     Hence,  by  Corollary  III,  p.  350,  we  have 

Theorem  II.    The  locus  of  two  equations  of  the  first  degree^ 

^     ^  \A^x^B^y-\-C^z  +  D^  =  0, 

is  a  straight  line  unless  the  coefficients  ofx,  y,  and  z  are  proportional. 

To  plot  a  straight  line  we  need  to  know  only  the  coordinates  of  two  points  on 
the  line.  The  easiest  points  to  obtain  are  usually  those  lying  in  the  coordinate 
planes,  which  we  get  by  setting  one  of  the  variables  equal  to  zero  and  solving  for 
the  other  two.  If  a  line  cuts  but  one  of  the  coordinate  planes,  we  get  only  one  point 
in  this  way,  and  to  plot  the  line  we  draw  a  line  through  that  point  parallel  to  the 
axis  which  is  perpendicular  to  that  plane. 

The  direction  of  a  line  is  known  when  its  direction  cosines  are 

known.     The  method  of  obtaining  these  is  illustrated  in 

Ex.  1.    Find  the  direction  cosines  of  the  line  whose  equations  are 

3x  +  2y-2;-l=:0,     2x-y  +  22;-3  =  0. 

Solution.    Let  the  direction  cosines  of  the  line  be  cos  a,  cos  j9,  and  cos  y. 
The  direction  cosines  of  the  normals  to  the  planes  in  which  the  line  lies 
are  respectively  (Corollary  I,  p.  350) 

3  2        _  _^  ^^^2     _  1     2 

Vli'   Vli         VTi  3         3    3 

363 


364  ANALYTIC  GEOMETRY 

Since  the  intersection  of  the  two  planes  is  perpendicular  to  the  normals  to 
both,  we  have  (Theorem  VI,  p.  335) 

o  2  1  2  12 

— izz  COS  a  +  —^=  cos  j3 =z  cos  7  =  0,     -  cos  (^^  —  ~  cos  i3  +  -  cos  7  =  0. 

Vl4  Vl4  Vl4  3  3  3 

Solving  for  cos  /3  and  cos  7  in  terms  of  cos  a,  we  get 

cos  j8  =  —  I  cos  a,     cos  7  =  —  |  cos  a, 

3cos/3          3  cos  7 
and  hence  cos  a  = —  = —  • 

Dividing  by  3,  the  least  common  multiple  of  the  numerators,  we  get 

cos  a  _  cos  /3  _  cos  7 

3     ~   -  8  ~   -7  * 

Then  by  the  Corollary,  p.  331, 

3^-8  -7 

cos  p  = ; —  J  cos  7 


±Vl22  ±Vl22  ±Vl22 

The  line  will  be  directed  downward  or  upward  according  as  the  positive 
or  negative  sign  of  the  radical  is  chosen. 

The  method  is  general  and  may  be  formulated  as  the 

Rule  to  find  the  direction  cosines  of  a  line  whose  equations  are  given. 

First  step.  Find  the  direction  cosines  of  the  7iormals  to  the  planes 
in  which  the  line  lies  (^Corollary  I,  p.  350). 

Second  step.  Find  the  conditions  that  the  given  line  is  perpen- 
dicular to  the  normals  in  the  first  step  (^Theorem  VI,  p.  S35)  and 
solve  for  two  of  the  direction  cosines  of  the  line  in  terms  of  the  third. 

Third  step.  Express  the  results  of  the  third  step  as  a  continued 
proportion  and  apply  the  Corollary,  p.  331. 

Ex.  2.    Find  the  direction  cosines  of  the  line  whose  equations  are 

4x  +  3z-10  =  0,     4a;-2y +  3z-l  =  0. 
Solution.    First  step.    The  direction. cosines  of  the  normals  to  the  given 
planes  are  4  3,4  2  3 

0,  -  and  — =,    --=^  —=■ 
5  5  V29  V29     V29 

Second  step.  If  the  direction  cosines  of  the  line  are  cos  a,  cos  jS,  and  cos  7, 
then 

-  cos  a  +  -  cos  7  =  0,     — — :  cos  a —=.  cos  p  -\ — ~=  cos  7  =  0, 

5  5  V29  ■V29  V29 

and  hence  cos  7  =  —  |  cos  a,     cos  ^  =  0. 


THE  STRAIGHT  LINE  IN  SPACE  365 

COS  Ct         COS  'V 

Third  step.    From  these  equations  — — -  = ,  cos  /3  =  0,  and  hence 

3  ■        —  4 
cos  a,  cos  13,  and  cos  7  are  proportional  to  3,  0,  and  —4.    Then  (Corollary, 

P- ^'^^^'  cosa:  =  ±|,    cosi8  =  0,    cos7  =  :f|. 

The  line  will  be  directed  downward  or  upward  according  as  the  upper  or 
lower  signs  are  used. 

Theorem  III.    If  a,  ^,  and  y  are  the  direction  cosines  of  the  line 
{II),  then 

cos  a  cos  p  cos  y 

This  is  proved  by  the  ahove  Rule  without  carrying  out  the  last  part  of  the  third 
step. 

PROBLEMS 

1.  Find  the  points  in  which  the  following  lines  pierce  the  coordinate  planes 
and  construct  the  lines. 

(a)  2a:  +  2/-z  =  2,  x-y  +  2z  =  4.        (c)  x  +  2y  =  8,  2a;-4y  =  7. 

(b)  4x  +  3?/-62  =  12,  4x-3y  =  2.      (d)  y  +  z  =  4,  x  -  ?/ +  2z  =  10. 

2.  Find  the  direction  cosines  of  the  following  lines. 

(a)  2x-2/  +  2z  =  0,  x+ 22/-22;  =  4.  _ 

Ans.    ±  6^3  V65,  T  g'^V^,  T  rV^65. 

(b)  x  +  2/  +  z  =  5,  x-2/  +  z  =  3.  Ans.    ±  i  V2,  0,  T  i  ^2. 

(c)  3x  +  2y-z  =  4,  x-2y-22;  =  5.    Ans.    ± /^ V5,  :f_l  V5,  ±  ^s^ V5. 

(d)  X  +  2/  -  32;  =  6,  2 X  -  2/  +  Sz  =  3.      Ans.    0,  ±  tV  VlO,  ±  yV  VlO. 

(e)  X  +  ?/  =  6,  2 X  -  3 z  =  5.  Ans.    ±  ^V V22,  T  ^h^/^,  ±  jV^^- 

(f)  y  +  3z  =  4,  32/-5z  =  l.  Ans.    ±1,0,0." 

(g)  2x-3?/  +  z  =  0,  2x-3i/-2z  =  6.  _ 

Ans.    ±  x^a  Vl3,  ±  fV  Vi3,  0. 
(h)  5x- 14z- 7  =0,  2x  +  7z  =  19.     ^ns.    0,  ±  1,  0. 

3.  Show  that  the  following  pairs  of  lines  are  parallel  and  construct  the 
lines. 

(a)  2?/  +  2:  =  0,  3?/  -  4z  =  7  and  5?/  -  2z  =  8,  4?/  +  11  z  =  44. 

(b)  x  +  2?/-z  =  7,  2/  +  2;-  2x  =  6  and  3x  +  6?/ -  3z  =  8,  2x- ?/ -  z  =  0, 

(c)  3x  +  z  =  4,  2/  +  2z=:9  and  6x-2/  =  7,  3?/+.6z  =  l. 

4.  Show  that  the  following  pairs  of  lines  meet  in  a  point  and   are 
perpendicular. 

(a)  x4-22/  =  l,  2y  —  z  =  l  and  x  —  y  =  l,x  —  2z  =  3. 

(b)  4x-fy-3z  +  24  =  0,  z  =  5  and  x  +  y  +  3  =  0,  x  +  2  =  0. 

(c)  3x  +  ?/-z  =  l,  2x-z  =  2  and  2x-2/+2z  =  4,  x-?/  +  2z  =  3. 


866 


ANALYTIC  GEOMETRY 


6,  Find  the  angles  between  the  following  lines,  assuming  that  they  are 
directed  upward  or  in  front  of  the  ZJT-plane. 


(a)  X 

(b)  X 


y  —  z  =  0,  y-^z  =  0  and  x  —  y  =  l,  x  — Sy  +  z  =  0.      Ans. 
2y  +  2z  =  l,x-2z  =  l  and  4x  +  3y 


z-\-l=0,2x+Sy  =  0. 
Ans.   cos-i^f. 
(c)  x-2y  -\-z  =  2,2y-z  =  l  Siud  x  -  2y  -^  z'=  2,  x  -  2y  +  2  z  =  4. 

Ans.    cos-ii. 

6.  Find  the  equations  of  the  planes  through  the  line 

x-\-y  —  z  =  0,  2x  —  y-\-Sz  =  6 
which  are  perpendicular  to  the  coordinate  planes. 

Ans.    3x  +  2z  =  5,  3y^5z-{-6  =  0,  5x  +  2y  =  5. 

7.  Show  analytically  that  the  intersections  of  the  planes  x  —  2y  —  z  =  Z 
and  2x  —  4y  —  2z  =  6  with  the  plane  x  +  y  —  Sz  =  0  are  parallel  lines. 

8.  Verify  analytically  that  the  intersections  of  any  two  parallel  planes 
with  a  third  plane  are  parallel  lines. 

160.  The  projecting  planes  of  a  line.  The  three  planes  passing 
throngh  a  given  line  and  perpendicular  to  the  coordinate  planes 
are  called  the  projecting  planes  of  the  line. 

If  the  line  is  perpendicular  to  one  of 
the  coordinate  planes,  any  plane  con- 
taining the  line  is  perpendicular  to  that 
plane.  In  this  case  we  speak  of  but 
two  projecting  planes,  namely,  those 
drawn  through  the  line  perpendicular 
to  the  other  coordinate  planes. 

If  the  line  is  parallel  to  one  of  the 
coordinate  planes,  two  of  the  projecting 
~j^  planes  coincide. 

By  Theorem  VII,  p.  360,  the 
equation  of  any  plane  through 
the  line 

(1)  A^x  +  B^y  +  C^z  4-  Z)i  =  0,    A^x  -f  B^y  -^  C^z  +  D^  =  0 
has  the  form 

(2)  (Ai  +  kA^)  ^  +  (5i  +  kB^)  2/  4-  (Ci  +  A^Cs)  ^  +  (A  +  kD^)  =  0. 
If  (2)  is  to  be  perpendicular  to  the  ZF-plane,  z  =  0,  then 

(Corollary  IV,  p.  350)  C,  +  kC^  =  0,  whence  k  =  --^-     Substi- 
tuting in  (2)  and  reducing,  we  get 

(3)  (C1J2  -  C^A,)x  -  (B,C,  -  B,C{)7j  +  CiA  -  C2A  =  0. 


THE  STRAIGHT  LINE  IN  SPACE 


367 


Similarly,  if  (2)  is  perpendicular  to  the  YZ-  or  ZX-plane,  it 
becomes 

(4)  (yliZ^a  -  A^B^)  y  -  (C^A^  -  C^A{)  z  +  AJ)^  -  A^D,  =  0, 

(5)  (A,B^  -  A^B,)  X  -  (B,C^  -  B^C^)  z  -  (B^D^  -  B^D,)  =  0. 

Equations  (3),  (4),  and  (5)  are  the  equations  of  the  projecting 
planes  of  the  line  (1),  and  any  two  of 
them  may  be  used  as  the  equations 
of  the  line. 

If  ^1^2  -  A^Bi-^  0,  that  is,  if  the 
-^  line  is  not  parallel  to  the  ZF-plane 
(Theorem  III),  equations  (5)  and  (4) 
may  be  written  in  the  forms 

X  =  mz  -\-  a,     y  =  nz  -\-  b. 


n, 


/ 


If  A^B^  —  J 2^1  =  0  and  B^C^  —  B^C^^^  0,  that  is,  if  the  line  is 
parallel  to  the  ZT-plane  but 
is  not  parallel  to  the  F-axis, 
equations  (5)  and  (3)  may  be 
written  in  the  forms 

z  =  a,     ?/  =  mx  +  b. 

If  A^B^  —  A^Bi  =  0  and 
BiC^  —  B2Ci  =  0,  that  is,  if  the 
line  is  parallel  to  the  F-axis, 
equations  (4)  and  (3)  may  be 
written  in  the  forms 

z  =  a,     X  =  b. 
Hence  we  have 

Theorem  IV.  The  equations  of  a  line  which  pierces  the  XY-plane, 
or  which  is  parallel  to  the  XY-plane  but  not  to  the  Y-axis,  or 
which  is  parallel  to  the  Y-axis,  may  be  put  in  the  following  forms 
respectively : 

(oc  =  mz-\-a,      (z  =  a,  (  z  =  a, 

^      ^  \y=nz-\-b,      \y  =  rnoc-\-b,     \qc  =  b. 


368  ANALYTIC  GEOMETRY 

To  find  the  equations  of  the  projecting  planes  of  a  given  line  we  may  proceed 
as  above  by  considering  the  system  of  planes  which  pass  through  the  given  line 
(Theorem  VII,  p.  3G0)  and  determining  the  parameter  k  so  that  the  plane  shall  be 
perpendicular  to  each  of  the  coordinate  planes  in  turn.  These  equations  may 
also  be  found  by  eliminating  2;,  x,  and  ij  in  turn  from  the  equations  of  the  line. 

To  reduce  the  equations  of  a  given  line  to  one  of  the  forms  (IV)  we  solve  them 
for  X  and  y  in  terms  of  z.  If  there  is  no  solution  for  x  and  y  (Theorem  IV,  p.  90), 
we  solve  for  y  and  z.  Finally,  if  there  is  no  solution  for  y  and  z,  we  solve  them 
for  z  and  x. 

PROBLEMS 

1.  Find  the  equations  of  the  projecting  planes  of  the  following  lines. 

(a)  2x  +  y-2;  =  0,  x-2/  +  22;  =  3. 

Ans.    5x '+?/  =  3,  3x  +  2;  =  3,  3^-52  +  6-0. 

(b)  X  +  ?/  +  z  =  6,  X  -  ?/  -  2  z  =  2. 

Ans.    3x  +  y  =  14,  2x-z  =  8,  2y+3z  =  4. 

(c)  2x  +  2/-z  =  l,  x-?/  +  z  =  2.  Ans.   x  =  l,  2/-z  +  l  =  0. 
(d)x4-2/-4z  =  l,  2x  +  22/  +  z  =  0.       Ans.    9x  +  92/  =  l,  9z  +  2  =  0. 

(e)  2 2/  +  3 z  =  6,  2 1/  -  3 z  =  18.  Ans.    y  =  Q,  z=-2. 

(f)  2x-2/  +  z  =  0,  4x  +  3y +  2z  =  6.    Ans.    5?/ =  6,  lOx  +  5z  =  C. 

(g)  X  +  z  =  1,  X  —  z  =  3.  Ans.    x  =  2,  z  =  —  1. 

2.  Reduce  the  equations  of  the  following  lines  to  one  of  the  forms  (IV) 
and  construct  the  lines. 

(a)  x  +  y  —  2z  =  0,  X  —  ?/  +  z  =  4.  Ans.  x  =  i z  +  2,  y  =  | z  —  2. 

(b)x  +  2y-z  =  2,  2x  +  42/  +  2z  =  5.  Ans.  z  =  i,  y  =  -  ^  x  +  |. 

(c)  X  —  2?/  +  z  =  4,  x  +  2?/  —  z  =  6.  Ans.  x  =  5,  y  =  iz-\-^. 

(d)  x  +  3z  =  6,  2x  + 5z  =  8.  Ans.  z  =4,x=-6. 
(e)x  +  22/  — 2z  =  2,  2x  +  ?/  — 4z=:l.  Ans.  x  =  2z,y  =  l. 
(f)  x-2/  +  z  =  3,  3x-3?/  +  2z=:6.  Ans.  z  =  S,  y  =  x. 

3.  Interpret  geometrically  the  meaning  of  the  constants  in  each  of  equa- 
tions (IV)  by  determining  numbers  proportional  to  the  direction  cosines  of 
each  line  and  the  point  in  which  the  first  line  cuts  the  JTY-plane,  the  second 
the  FZ-plane,  and  the  third  the  ZX-plane. 

4.  Interpret  the  geometric  significance  of  the  constants  in  equations  (IV) 
by  considering  the  traces  of  the  planes  which  are  the  loci  of  those  equations 
taken  separately. 

5.  Show  that  a  straight  line  in  space  is  determined  by  four  conditions,  and 
formulate  a  rule  by  which  to  find  its  equations. 

6.  Find  the  equations  of  the  line  passing  through  the  points  (—2,  2,  1) 
and  (-  8,  5,  -  2).  Ans.   x  —  2  z  -  4,  y  =  —  z  -\-  S. 


THE  STRAIGHT  LINE  IN  SPACE  369 

7.  Find  the  equations  of  the  projection  of  the  line  a;  =  z  +  2,  2/  =  2z  —  4 
upon  the  plane  x  +  y  —  z  =  0.  Ans.    x  =  ^  z  +  Y,  2/  =  1 2;  —  Y-. 

8.  Find  the  equations  of  the  projection  of  the  line  z  =  2,  y  =  x  —  2  upon 
the  plane  X  —  2y  —  3z  =  4.  Ans.    x=:— 6z  +  4,  y  =  — 4^;. 

9.  Show  that  the  equations  of  a  line  may  be  written  in  one  of  the  forms 

fy  =  mx  +  a,       fx  =  a,  Jx  —  a, 

\z=  nx  +b,       \z  =  my  -]-b,      \y  =  b, 

according  as  it  pierces  the  FZ-plane,  is  parallel  to  the  FZ-plane,  or  is  parallel 
to  the  Z-axis. 

10.  Show  that  the  condition  that  the  line  x  =  mz  -\-  a,  y  =  nz  -\- b  should 

,     ,.  ,         ,  ,       ^,  .     a  —  a'        b  —b' 

mtersect  the  Ime  x  —  m'z  +  a,  y  =  nz  +  0'  is = -• 

m  —  m       n  —  n 

161.  Various  forms  of  the  equations  of  a  straight  line. 

Theorem  V.  Parametric  form.  The  coordinates  of  any  point 
P(x,  1/,  z)  on  the  line  through  a  given  point  Pxioc\,  yi,  z-^  whose 
direction  angles  are  a,  /3,  and  y  are  given  by 

(V)    oc  =  oci-\-  p  cos  a,    y  =  y^-\-  pcos  p,    z  =  z^  +  p  cos  y, 

where  p  denotes  the  variable  directed  length  PiP. 

Proof.  The  projections  of  PiP  on  the  axes  are  respectively 
(Corollary  II,  p.  329) 

x-x^,      y-yi,      « -  «i, 

or  (Theorem  I,  p.  328) 

p  cos  a,     p  COS  /3,     p  cos  y. 
Hence 

X  —  Xi  =  p  cos  ct,     y  —  yi  =  p  cos  p,     z  —  Zi  =  p  cos  y. 

Solving  for  x,  y,  and  z,  we  obtain  (Y).  q.e.d. 

Theorem  VI.  Symmetric  form.  The  equations  of  the  line  passing 
through  the  point  P^  (xi,  y^,  z^)  whose  direction  angles  are  a,  ^,  and 
y  have  the  form 

^     ^  cos  a         cos  p        cos  y 

Hint.   Solve  each  of  equations  (V)  for  p  and  equate  the  values  obtained. 


370  ANALYTIC  GEOMETRY 

^      ,,  ^.  cosa       cos/8      cosy 

Corollary.    If =  — y~-  — -y  then  the  sijmmetnc  equations 

of  the  line  may  he  written  in  the  form 

X  —  Xj_      y  —  Vi      z  —  Zi 


a  h  c 

Theorem  VII.  Two-point  form.  The  equations  of  the  straight  line 
passing  through  Pi  (iCj,  y^,  z^  and  Pg  (^2?  2/2?  --2)  <^^^ 

^        ^  i»2  —  «i      2/2  —  2/1      «2  — «i" 

Froof.  The  line  (VI)  passes  through.  Pj.  If  it  also  passes 
through  P2,  then 

cos  a  cos  ^  COS  y 

Dividing  (VI)  by  this  equation,  we  obtain  (VII).  q.e.d. 

Equations  (VI)  and  (VII)  each  involve  three  equations,  namely,  those  obtained 
by  neglecting  in  turn  each  of  the  three  ratios.  These  equations  are,  in  different 
form,  the  equations  of  the  projecting  planes,  since  one  variable  is  lacking  in  each 
(Corollary  V,  p.  351).  Any  two  of  the  three  equations  are  independent  and  may 
be  used  as  the  equations  of  the  line,  but  all  three  are  usually  retained  for  the  sake 
of  their  symmetry. 

PROBLEMS 

1.  Find  the  equations  of  the  lines  which  pass  through  the  following  pairs 
of  points,  reduce  them  to  one  of  the  forms  (IV),  p.  367,  and  construct  the 
lines. 

(a)  (3,  2,  -  1),  (2,  -  3,  4).  Ans.  x  =  -\z^  V,  2/  =  -  2  +  1- 

(b)  (1,  6,  3),  (3,  2,  3).*  An».  2=^3,  y  =  -2x  +  8. 

(c)  (1,  -  4,  2),  (3,  0,  3).  Ans.  x  =  2z  -  S,  y  =  4z  -  12. 

(d)  (2,  -  2,  -  1),  (3,  1,  -  1).  Ans.  z  =  -l,y  =  Sx-8. 

(e)  (2,  3,' 5),  (2,  -  7,  5).  Ans.  z  =  6,x  =  2. 

2.  Show  that  the   two-point  form  of  the  equations  of  a  line  become 

^~^^  =  y  ~y^  ,  z  =  Zi,  if  zi  =  Z2.     What  do  they  become  if  2/1  =  2/2? 
X2  -xi      2/2  -  2/1 

if  xi  =  X2  ? 

*  From  (VII),  '^^  =  ^-^  =  ^-^-  •   The  value  of  the  last  ratio  is  infinite  unless  2-3=0. 
3-1      2-6      3-3 
If  3-3  =  0,  then  the  last  ratio  may  have  any  value  and  may  be  equal  to  the  first  two. 

Hence  the  equations  of  the  line  become  ^-^  = ,  z  =  3.    Geometrically,  it  is  evident 

2  -4 

that  the  two  points  lie  in  the  plane  z  =  3,  and  hence  the  line  joining  them  also  lies  in  that 
plane. 


THE  STRAIGHT  LINE  IN  SPACE  371 

3.  What  do  the  two-point  equations  of  a  line  become  if  Xi  =  X2  and 
2/1  =  2/2?   if  yi  =  2/2  and  zi  =  Z2?   if  Zi  =  Z2  and  xi  —  x^? 

4.  Do  the  following  sets  of  points  lie  on  straight  lines  ? 

(a)  (3,  2,  -  4),  (5,  4,  -  6),  and  (9,  8,  -  10).  Ans.    Yes. 

(b)  (3,  0,  1),  (0,  -  3,  2),  and  (6,  3,  0).  Ans.    Yes. 

(c)  (2,  5,  7),  (-  3,  8,  1),  and  (0,  0,  3).  Ans.   No. 

5.  Show  that  the  conditions  that  the  three  points  Pi (iCi.yi,  21),  P2(iC2, 2/2, 2^2), 

and  Ps  (X3, 2/3,  23)  should  lie  on  a  straight  line  are  ^LH^  =  Vs  -  Vi  ^  Zs  -  Zi 

«2  -xi      2/2  -  2/1      22  -  2l 

6.  Find  the  equations  of  the  line  passing  through  the  point  (2,  -1,  -  3) 
whose  direction  cosines  are  proportional  to  3,  2,  and  7,  and  reduce  them  to 
the  form  (IV),  p.  367.  Ans.   x  =  f  2  +  V,  y  =  ^z-  ^. 

7.  Find  the  equations  of  the  line  passing  through  the  point  (0,  —  3,  2) 
which  is  parallel  to  the  line  joining  the  points  (3,  4,  7)  and  (2,  7,  5). 

X      2/  +  3      z  -2 


Ans. 


1-3  2 


8.  Show  that  the  lines  ^^  =  ^^^  ..  "  and  ^±i  =  ^^  =  ^±^  are 
parallel.  ^  -2        4  -3  2  -4 

9.  Find  the  equations  of  the  line  through  the  point  (—2,  4,  0)  which  is 

jjj      ■?/  -I-  2      z 4 

parallel  to  the  line  -  = = ,  and  reduce  them  to  the  form  (IV),  p.  367. 

^  ~^         Ans.    x  =  -iz-2,y  =  -3z  +  4. 

■tn    ax,       *u^^i,T       x  +  2      y  —  S      z  —  1       ,x  — 3      y      z  +  3 

10.  Show  that  the  hues  — —  =  - — —  = and =  -^  =  — ■ —  are 

perpendicular.  ^  ~^  ^  2  6  3 

11.  Find    the    angle    between    the    lines    =  - —      ~       and 

2  1-1 

cc  +  22/-72;.-,^,  J.      ,j  ^  ^        27r 

— :; —  = =  -  if  both  are  directed  upward.  Ans. 

12  1  ^  •  3 

12.  Find  the  parametric  equations  of  the  line  passing  through  the  point 
(2,  —  3,  4)  whose  direction  cosines  are  proportional  to  1,  —  2,  and  2. 

Ans.    x  =  2  +  ip,  2/  =  -  3  -  f /),  2  =  4  +  f  p. 

13.  Construct  the  lines  whose  parametric  equations  are 

(a)  X  =  2  +  fp,  2/  =  4  -  ip,  z  =  6  +  f  p. 

(b)  X  =  -  3  -  f /),  ?/ =  6  -  f  p,  z  =  4  +  f  p. 

14.  Find  the  distance,  measured  along  the  line  x  =  2  —  ^gp,  2/  =  4  +  j|/), 
2  =  —  3  +  j-**3  p,  from  the  point  (2,  4,  —  3)  to  the  intersection  of  the  line  with 
the  plane  4x  —  2/  —  2z  =  6.  Ans.    If. 

16.  Show  that  the  symmetric  equations  of  the  straight  line  become 

^  =  - — — »  2  =  2i  if  cos  7  =  0.     What  do  they  become  if  cos  a  =  0  ? 

cos  or        cos/3 

if  cos/3  =  0? 


372  ANALYTIC  GEOMETRY 

16.  Show  that  the  symmetric  equations  of  the  straight  line  become  z  =  zi, 
x  =  Xi  if  cos  7  =  cos  a  =  0.  What  do  they  become  if  cos  a  =  cos  j3  =  0  ? 
if  cos  /3  =  cos  7=0? 

17.  Reduce  the  equations  of  the  following  lines  to  the  symmetric  form. 

(a)  x-2y +  z  =  8,  2x-3?/  =  13.  Ans.   ^^I^  =  ^-±-?  =  - . 

O  JJ  1. 

(b)  4x-5?/  +  3z  =  3,  4x-5y  +  z  +  9  =  0.     Ans.    -  =  ^-^ ,  z  =  Q. 

5  4 

(c)  2x  +  z  +  5  =  0,  x  +  32;-5=0.  Ans.   2  =  3,  x  =  -  4. 

(d)x  +  22/  +  62  =  5,  3x-2y-10z  =  7.  Ans.   ?-^l- =  ^^^  =  1 

2  —  7        2 

/;p  3        2 

(e)3x-2/-22;  =  0,  6x-3?/-42;  +  9  =  0.     ^ns.    =  -,2/ =  9. 

2  o 

(f)  3  X  -  4  y  =  7,  X  +  3  2/  =  11.  ^ns.    x  =  5,  y  =  2. 

(g)  2x  +  y  +  22  =  7,  x  +  3y+62;  =  ll.  Ans.   ^— —  =  — - ,  x  =  2. 

A  —  i 

(h)2x-3?/  +  z  =  4,  4x-6y-z  =  5.  Ans.    -=y^^,  z  =1. 

o  2 

(i)  3  2  +  ?/  =  1,  4  2;  -  3  y  =  10.  Ans.    y  =  -2,z  =  l. 

^        X  —  a      y  —  h      z 

(])  X  =  mz  -}-  a,  y  =  nz  -{-  b.  Ans. = =  -■ 

m  n  1 

Hint.  Find  the  coordinates  of  a  point  on  the  line  hy  assuming  a  value  of  one  variahl* 
and  solving  the  equations  of  the  line  for  the  other  two  variables.  In  the  answers  thi 
point  is  the  point  in  which  the  line  pierces  the  X  T-plane,  or  the  point  in  which  it  pierce 
the  FZ-plane  if  it  is  parallel  to  the  XF-plane,  or  the  point  in  which  it  pierces  the  ZX- 
plane  if  it  is  parallel  to  the  T-axis. 

Find  the  direction  cosines  of  the  line  by  the  Rule,  p.  364  (or  numbers  proportional  to 
them  by  Theorem  III,  p.  365),  and  substitute  in  the  symmetric  equations  of  the  line  (oi 
in  the  form  given  in  the  Corollary  to  Theorem  VI). 

If  one  or  two  of  the  direction  cosines  are  zero,  the  symmetric  equations  take  the  fori 
given  in  problems  15  and  16. 

18.  Find  the  equations  of  the  line  passing  through  the  point  ^2, 0,  —2)  which 

is  perpendicular  to  each  of  the  lines =  -  =  and  -  = — -  =  — — 

2  1  2  o        —  1  2 

Ans.   l^  =  ?^  =  i±?. 
4  2        -5 

19.  Find  the  equations  of  the  line  passing  through  the  point  (3,  —1,2)  which 

Ls  perpendicular  to  each  of  the  lines  x  =  2z  —  1,  y  =  z  +  B,  and  -  =  -  =  -. 

^       o       4 

.        x-3      y+1      z-2 
Ans.   — -— = —  =  — - — 


THE  STRAIGHT  LINE  IN  SPACE  373 

20.  Find  the  equations  of  the  line  through  Pi(xi,  2/1,  Zi)  parallel  to 

,  .  x-Xi     y  -Vi     Z-Z2 

(a)  =  — 7 — -  = 

a  b  c 

(b)  X  =  mz  +  a,  y  =  nz  +  h. 


(c)  z  =  a,  y  =  mx  +  b. 

(d)  Aix  +  Biy  +  Ciz  +  Di  =  0,  A^x  +  B^y  +  C^z 

A  X-Xi 

Ans. 


A            X-Xi 

Ans.   -  = 

a 

y. 

-yi 

b 

Z-Zi 

c 

.            X  —  X\ 

Ans.   i  = 

m 

y. 

-yi 

n 

Z-Zx 

1 

X  —  Xi 

Ans. = 

1 

L 

~yi, 

m 

Z  =  Zi. 

B^y  +  C2Z  +  A 

J  = 

0. 

y-yi 

z 

-zx 

BIC2  —  B2CI        CIA2  —  A2CI       AIB2  —  A2BI 
21.  Find  the  equations  of  the  line  passing  through  Pi(Xi,  2/1,  Zi)  which  is 
perpendicular  to  each  of  the  lines 

X  —  X2  _  y  —  z/2  _  z  -  Z2      ,  x-xs  _  y  -  ys  _  z  -  zz 

a%  &2  C2  Ct3  63  C3 

-xx  y  -y\  z-zi 


Ans. 


&2«3  —  bsCz         Citts  —  Cza2         ^263  —  «3&2 

162.  Relative  positions  of  a  Hne  and  plane.  If  the  equations 
of  a  line  have  the  general  form  (II),  p.  363,  then  the  line  will  lie 
in  a  given  plane  if  a  value  of  k  in  (VII),  p.  360,  may  be  found 
such  that  the  locus  of  that  equation  is  the  given  plane. 

If  the  equations  of  the  line  have  the  form  (IV),  we  substitute 
the  values  of  two  of  the  variables  given  by  (IV)  in  the  equation 
of  the  plane  and  see  whether  the  result  is  true  for  all  values  of 
the  third  variable.    If  such  is  the  case,  the  line  lies  in  the  plane. 

An  analogous  procedure  may  be  followed  if  the  equations  of 
the  line  have  the  form  (V),  (VI),  or  (VII). 

Theorem  VIII.   A  line  ivhose  direction  angles  are  a,  ^,  and  y  and 
the  plane  Ax  -{-  By  -\-  Cz  -\-  D  =  0  are 
(a)  parallel  when  and  only  when 

Acqs  a  + B  cos  p -\- C  cosy  =  0\ 
(h)  perpendicular  when  and  only  when 
ABC 
cos  a  ~  COS  jff  ~  cos  y 
Proof.    The  direction  cosines  of  the  normal  to  the  plane  are 
(Corollary  I,  p.  350) 

A B C 


374  ANALYTIC  GEOMETRY 

The  line  and  plane  are  parallel  when  and  only  when  the  line 
is  perpendicular  to  the  normal  to  the  plane, "*  that  is  (Theorem  VI, 
p.  335),  when  and  only  when 

A  Gos  a  -{-  Bcos  13  +  C  cos  y  _ 

Multiplying  by  the  radical,  we  get  the  condition  for  parallelism. 

The  line  and  plane  are  perpendicular  when  and  only  when 
the  line  is  parallel  to  the  normal  to  the  plane,  that  is  (Theorem  VI, 
p.  335),  when  and  only  when 

cos  a  =  — — ,      COS  /?  = 


±V^2  +  i>''  +  C'  ±-^A-'  +  B-'  +  C^ 

C 


cos  V  = 


±  Va^  -^B^  +  C^ 

Dividing  these  equations  by  A,  B,  and  C  respectively  and 
inverting,  we  at  once  obtain  the  conditions  for  perpendicularity. 

Q.E.D. 

163.  Geometric  interpretation  of  the  solution  of  three  equa- 
tions of  the  first  degree.  The  coordinates  of  a  point  which  lies 
on  each  of  three  planes  will  satisfy  the  equations  of  the  three 
planes,  and  hence  to  each  point  common  to  three  planes  there 
will  correspond  a  solution  of  their  equations.  Hence  we  have 
the  following  correspondence  between  the  relative  positions  of 
three  planes  and  the  number  of  solutions  of  their  equations. 

Position  of  planes  Number  of  solutions  of  equations 

forming  a  triedral  angle.  One  solution. 

Forming  a  prismatic  surface,  t       No  solution. 
Passing  through  the  same  line.$     A  singly  infinite  number.  § 
Three  parallel  planes.  $  No  solution. 

Three  coincident  planes.  A  doubly  infinite  number.  || 

*  If  the  line  is  perpendicular  to  the  normal  to  the  plane,  it  may,  in  a  special  case,  lie 
in  the  plane. 

t  Two  of  the  planes  may  be  parallel  in  a  special  case. 
X  Two  of  the  planes  may  coincide  in  a  special  case. 
§  The  solution  contains  one  arbitrary  constant. 
II  The  solution  contains  iwo  arbitrary  constants. 


THE  STRAIGHT  LINE  IN  SPACE  375 

If  the  three  planes  form  a  triedral  angle,  the  point  of  intersection  is  found 
without  difficulty  by  solving  their  equations. 

If  the  three  planes  form  a  prismatic  surface,  their  lines  of  intersection  are  par- 
allel. Whether  this  is  the  case  or  not  may  be  determined  by  Theorem  III,  p.  365, 
and  the  Corollary,  p.  331. 

If  the  three  planes  pass  through  the  same  line,  the  intersection  of  two  planes 
lies  in  the  third.  Whether  this  is  the  case  or  not  may  be  determined  by  the 
method  on  p.  373.  To  solve  their  equations  set  one  variable  equal  to  k  and  solve 
tioo  of  the  equations  for  the  remaining  variables.  The  results  will  be  solutions 
for  all  values  of  k. 

Whether  the  three  planes  are  parallel  or  not  may  be  determined  by  Corollary  III, 
p.  350. 

If  the  three  planes  coincide,  all  of  their  coefficients  are  proportional.  To  solve 
their  equations  set  two  of  the  variables  equal  to  k\  and  k-z,  and  solve  one  of  the 
equations  for  the  remaining  variable.  The  results  will  be  solutions  for  all  values 
of  k\  and  ki. 

PROBLEMS 

jc  _j.  3       y 4      z 

1 .  Show  that  the  line = =  -  is  parallel  to  the  plane  4  x  +  2  y 

+  22  =  9.  ^  -7        3 

2.  Show  that  the  line  -  =  -  =  -  is  perpendicular  to  the  plane  3x  +  2?/ 
+  7z  =  8.  3      2      7 

3.  Show  that  the  line  x  =  2;  —  4,  2/  =  2z— 3  lies  in  the  plane  2x  —  Zy 
+  42-1  =  0. 

4.  Show  that  the  line  x=-2  +  f/3,  y  =  -|p,  z  =  6  +  ip  lies  in  the  plane 
x-2y-6z  +  38  =  0. 

5.  Find  the  coordinates  of  the  points  of  intersection  of  the  following 
planes  and  determine  the  relative  positions  of  the  planes. 

(a)  2x  +  y-22;  =  ll,x-?/  +  2;  =  0,  jc  +  2?/-2;  =  7. 

Ans.    (3,  1,  —  2);  planes  form  a  triedral  angle. 
(b)2x  +  42/  +  22  =  3,  3x  +  3?/  +  2;  =  0,  3x-6y-52;==8. 

Ans.    None;  planes  form  a  prismatic  surface, 
(c)  X  —  ?/  —  32  =  1,  x  +  y  +  2;=:2,  3x  —  y  —  5-2;  =  4. 

Ans.    (I  +  A:,  I  —  2 fc,  A:);  planes  pass  through  a  line. 
(d)3x-2/  +  52;  =  0,  21x-7y  +  35z  =  8,  22/-10z-6x  =  4. 

Ans.   None ;  planes  are  parallel. 
(e)2x-3y  +  42  =  3,  6y-4x-8z  +  6  =  0,  6x-9y  +  12 z  =  9. 

Ans.    [fci,  ^2,  1(3  —  2  A;i  +  3  ^2)] ;  planes  coincide. 

6.  Show  that  the  line  ^-^-  =  ^ =  — - —  lies  in  the  plane  2  x  +  2  2/ 

-2  +  3  =  0.  ^  -^  ^ 


376  ANALYTIC  GEOMETRY 

7.  Find  the  equations  of  the  line  passing  through  the  point  (3,  2,  —  6) 

which  is  perpendicular  to  the  plane  ix  —  y  +  Sz  =  d. 

x-3      y -2      2+6 

Ans.    = = 

4-13 

8.  Find  the  equations  of  the  line  passing  through  the  point  (4,  —  6,  2) 
which  is  perpendicular  to  the  plane  x-{-  2y  —  Sz  =  8. 

.         ic-4      y  +  6      2-2 
1  2  -3 

9.  Find  the  equations  of  the  line  passing  through  the  point  (—2,  3,  2) 
which  is  parallel  to  each  of  the  planes  Bx  —  y  -i-  z  =  0  and  x  —  z  =  0. 

x  +  2      y - S      z -  2 


Ans. 


1  4  1 


10.  Find  the  equation  of  the  plane  passing  through  the  point  (1,3,  —  2) 
which  is  perpendicular  to  the  line = = 

Ans.    2x  +  5y  —  z  =  19. 

11.  Find  the  equation  of  the  plane  passing  through  the  point  (2,  —  2,  0) 
which  is  perpendicular  to  the  line  z  =  3,  y  =  2x  —  i.     Ans.  x-\-2y  +2  =  0. 

12.  Find  the  equation  of  the  plane  passing  through  the  line  x  +  2  z  =  4, 

3j 3      ■j/_l4      2; 7 

y  —  z  =  8  which  is  parallel  to  the  line = = . 

2  3  4 

Ans.    x  +  lOy -8z-84:  =  0. 

13.  Find  the  equation  of  the  plane  passing  through  the  point  (3,  6,  —  12) 

which  is  parallel  to  each  of  the   lines  — —  =  ^  ~     =  ^-^   and  ^  ~ 
_z+2  3-13  2 

-—^'2/ -3.  Ans.    2x-\-3y  -z  =  S6. 

14.  Find  the  equations  of  the  line  passing  through  the  point  (3,  1,  —  2) 
which  is  perpendicular  to  the  plane  2x  —  y  —  5z  =  6. 

Ans.   x  =  -|z  +  -U,  2/  =  i2;  +  |. 

15.  Show  that  the  lines  ^^  ^  ^±i  ^  ^     ^^d     ^^  =  ^±i  ^  ^ 

3  4-2  -1  32 

intersect,  and  find  the  equation  of  the  plane  determined  by  them. 

Ans.    14  X  -  4  2/  +  13  z  =  32. 

/>.  2         y    I    3 

16.  Find  the  equation  of  the  plane  determined  by  the  line  =: 

z-1  2-2 


1 


and  the  point  (0,  3,  -  4).  Ans.   x-^2y  +  2z-^2  =  0. 


17.  Find   the   equation  of  the   plane   determined  by  the  parallel   lines 
x-^l  _y  -2  _z  X  -S  _y  +  4  _z  -1 

3  2  13  2  1        Ans.    8x  +  y-26z-\-6  =  0. 


THE  STRAIGHT  LINE  IN  SPACE  377 

18.  Find  the  equations  of  the  line  passing  through  Pi(xi,  ?/i,  Zi)  which  is 
perpendicular  to  the  plane  Ax  -{-  By  i-  Cz  +  D  =  0. 

Ans.   ^^1^  =  2^1-1  =  ^^111. 
ABC 

19.  Find  the  equation  of  the  plane  passing  through  the  point  Pi  (cci,  yi,  zi) 

which  is  perpendicular  to  the  line =  — ; — -  = . 

a  0  c 

Ans.   a{x-Xi)-\-b{y-  yi)  -\-  c{z-Zi)  =  0. 

20.  Find  the  angle  6  between  the  line = —  = and  the 

plane  ^x  +  ^y  +  C2;  +  D  =  0.  "*        .     ^  .,.      ^^ 

,          .    ^                  Aa  +  Bb  +  Cc 
Ans.    sm  9  = — 

Hint.  The  angle  between  a  line  and  a  plane  is  the  acute  angle  between  the  line  and 
its  projection  on  the  plane.  This  angle  equals  ~  increased  or  decreased  by  the  angle 
between  the  line  and  the  normal  to  the  plane. 

21.  Find  the  equation  of  the  plane  passing  through  Ps{xs,  ys?  Zs)  which 

11  w        u    r  *i     T       x-xi     y  -y\     z-zi      .x-x^     v  -yi 

IS  parallel  to  each  of  the  lines  =  — - —  = and =  f__^ 

ai  &i  ci  ai  02 

Z  —  Z2 


Ans.   (61C2-&2C1)  {x-Xz)  +  {cia2-a2C\)  (2/-2/3)  +  (ai?>2-a2&i)  {z-Zz)=Q. 

22.  Find  the  condition  that  the  plane  A^x  +  B\y  +  C\Z  +  Di  =  0  should 
be  parallel  to  the  line  A^x  +  B^y  +  C^z  +  D2  =  0,  A^x  +  B^y  +  C^z  +1)3  =  0. 

Ans.     Ai{B2Cs  -^3C2)  +  5i(C2^3-C3^2)+Ci(^2^3-^3-B2)=  0. 

23.  Find  the  equation  of  the  plane  determined  by  the  point  Pi  {xi,  yi,  Zi) 
and  the  line  Aix  -{■  Biy  +  CiZ  +Di  =  0,  A2X  +B2y+  C^z  +D2  =  0. 

Ans.   [A^xi  +  B^yi  +  C^Zi  +D2)  {A-LX+B^y+  C^z  +i)i) 

=  (^ixi  +  jBi2/i  +  Cizi  +Di)  {A2X  +B2y  +  C^z  +D2). 

24.  Find  the  equation  of  the  plane  determined  by  the  intersecting  lines 
x-xi  _  y-yi  _  z  -zi  ^^^  x  -  xi  _  y  -  yi  _  z  -  zi  _ 

ai  6i  ci  a2  62  C2 

Ans.    (61C2  -  &2C1)  (x  - Xi)  +  (cia2  -  C2ai)  {y  -  yi)  +  (ai62 — aa&i)  {z  -  Zi)  =  0. 

25.  Find  the  equation  of  the  plane  determined  by  the  parallel  lines 
X  — xi  _y  —  yi  _z  —  zi      ,  x  —  X2  _  y  —  1/2  _  z  —  Z2  ^ 

a  b  c  a  b  c 

Ans.    [(2/1  -  2/2)  c  -  (zi  -  Z2)  b]x  +  [(zi  -  Z2)  a  -  (xi  -  X2)  c]  y 
+  [(xi  -  X2)  &  -  (^1  -  2/2)  a]  z  +  (yiZ2  -  y2Zi)  a 

+  (2:1X2  -  Z2X1)  b  -1-  (Xi?/2  -  iC22/l)  c  =  0. 


378  '  ANALYTIC  GEOMETRY 

26.  Find  the  conditions  that  the  line  x  =  mz  -{-  a,  y  =  nz  +h  should  lie  in 
the  plane  Ax  +  By  -\-  Cz  +  D  =  0. 

Ans.   Aa  +  Bb-\-D  =  0,  Am  +  Bn  +  C  =  0. 

27.  Find   the  equation   of  the  plane  passing  through   the  line 

6i  C\  a^  62  C2 

Ans.    (61C2  —  62C1)  (x — cci)  +  (cia2  —  C2ai)  (2/ — 2/1)  +  (ai&2  —  a2&i)  (2  —  Zi)  =  0. 


i;^ 


CHAPTER   XX 
SPECIAL  SURFACES 

164.  In  this  chapter  we  shall  consider  spheres,  cylinders,  and 
cones  ^  (surfaces  considered  in  Elementary  Geometry)  and  sur- 
faces which  may  be  generated  by  revolving  a  curve  about  one  of 
the  coordinate  axes  or  by  moving  a  straight  line. 

165.  The  sphere. 

Theorem  I.  The  equation  of  the  sphere  whose  center  is  the  point 
(tr,  /?,  y)  and  whose  radius  is  r  is 

(a^-ay  +  (v-l3y  +  (z-Yy  =  A  or 
(I)_  3c''-^y^  +  z^-2ax--  Q^y  -  2  ys;  +  a^  +)8'^  +  y^  -  r-^  =  O. 

Froof.  Let  P  (x,  y,  z)  be  any  point  on  the  sphere,  and  denote 
the  center  of  the  sphere  by  C.  Then,  by  definition,  PC  =  r. 
Substituting  the  value  of  PC  given  by  (IV),  p.  331,  and  squar- 
ing, we  obtain  (I).  q.e.d. 

Theorem  II.    The  locus  of  an  equation  of  the  form 
(II)  ic'-^tf  +  z'  +  GQC-\-  Hy  -^Iz  +  KzzzO 

is  determined  as  follows : 

(a)    When  G'^  -\-  H"^  -\-  P  —  ^  K  >  0,  the  locus  is  a  sphere  whose 

center  is  (  — —^  — ^j  "~  o  )  ^^^  whose  radius  is 


{h)    When  G'^  +  H^  +  I^  —  4:K=  Q,  the  locus  is  the point-spheret 
(g_    _H    _/ 
V     2'        2'       2, 

(c)    When  G^  +  H^ -^  I^  -  4:K  <  0,  there  is  no  locus. 

*  In  Analytic  Geometry  the  terms  sphere,  cylinder,  and  cone  are  usually  used  to 
denote  the  spherical  surface,  cylindrical  surface,  and  conical  surface  of  Elementary 
Geometry,  and  not  the  solids  bounded  wholly  or  in  part  by  such  surfaces. 

+  That  is,  a  point  or  sphere  of  radius  zero. 

379 


380  ANALYTIC  GEOMETRY 

Proof.    Comparing  (II)  with  (I),  we  obtain 

whence 


G       ^  H  I 


Hence,  if  G^  +  H^  -{■  I^  -  4:  K  >  0,  the  locus  is  a  sphere. 

To  determine  the  general  appearance  of  the  locus  of  (II)  when 
G^  -{-  H^  -\-  I^  —  4:K^0,  we  consider  the  section  formed  by  the 
plane  z  =  k,  whose  equation  is  (Eule,  p.  345) 

(1)  x''  +  2f-{-Gx-{-Hi/-^k''  +  Ik-^K=  0. 

The  discriminant  of  (1)  is  (p.  131) 

©  =  G'^  +  7/2  _  4  A;2  _  4  Ik  -  4  k'^^ 
=  -  4  7^2  -  4  //i:  +  G^2  ^  iy2  _  4  K 

The  discriminant  of  this  quadratic  in  k  is  (p.  2) 
A  =  16  /2  +  16  (9^  +  16  ^2  -  64  ii: 
=  16(G'2  +  //2_^/2_4  7r). 

In  discussing  the  locus  of  (1)  three  cases  arise  which  depend 
upon  the  sign  of  ©  (Theorem  I,  p.  131). 

(a)  If  (92  +  iy2  +  72  _  4  7^  >  0,  0  is  positive  for  values  of  k 
lying  between  the  roots  of  ©  (Theorem  III,  p.  11),  and  the 
section  (1)  formed  by  th,e  plane  ^  =  A;  is  a  circle.  Equation  (II) 
has  a  locus,  as  we  have  seen. 

(b)  If  (92  +  7^2  4-  72  -  4  7:  =  0,  ©  is  negative  for  all  real  values 
of  k  (Theorem  III,  p.  11)  except  the  roots,  which  are  real  and 
equal  (Theorem  II,  p.  3),  and  for  this  single  value  of  k  the  locus 
of  (1)  is  a  point-circle.  As  but  one  plane,  z  =  k,  intersects  the 
locus  of  (II),  and  as  this  intersection  is  a  point-circle,  the  locus 
is  a  point  which  may  be  regarded  as  a  sphere  of  zero  radius. 

(c)  If  (92  -f  TT^  +  72  -  4  A^  <  0,  ©  is  negative  for  all  real  values 
of  k  (Theorem  III,  p.  11).  Hence  (1)  has  no  locus  whatever  the 
value  of  k  may  be,  and  therefore  (II)  has  no  locus.  q.e.d. 


SPECIAL  SURFACES  381 

Theorem  III.    The  locus  of  the  general  equation  of  the  second 
degree  in  three  vanables 

(III)   Ax^-\-  Bi/  +  Cz^  +  Dgz  H-  Ezx  +  Fxij  +  Gx  -\-  Hg  -\-Iz  +  7i  =  0 

{5  a  sphere  when  and  only  when  A  =z  B  =  C,  D  =  E  =  F  =  Oj  and 

G^ +  !{'-{-  1^-4: AK   . 
—^ is  positive. 

This  is  proved  by  comparing  (III)  with  (II). 


PROBLEMS 

1 .  Find  the  equation  of  the  sphere  whose  center  is  the  point 

(a)  {a,  0,  0)  and  whose  radius  is  a.  Ans.   x^  -{-  y^  +  z^  —  2  ax  =  0. 

(b)  (0,  iS,  0)  and  whose  radius  is  /3.  Ans.   x^  -\- y^ -\-  z^  -  2  ^y  =0. 

(c)  (0,  0,  7)  and  whose  radius  is  7.  Ans.   x^  -{-  y^  +  z^  —  2yz  =0. 

2.  Determine  the  nature  of  the  loci  of  the  following  equations  and  find  the 
center  and  radius  if  the  locus  is  a  sphere,  or  the  coordinates  of  the  point- 
sphere  if  the  locus  is  a  point-sphere. 

(a)  x2  -f  2/^  -H  22  -  6 X  +  4 z  =  0.         (c)  x^  -{-  y"^  +  z^  +  ix  -  z  +  7  =  0. 

(b)  x2-|-2/2  +  22  +  2a;-4y-5  =  0.  (d)  x'^  -\-  y^ -{-  z^  -  12x  +  6y  +  4: z  =  0. 

3.  Where  will  the  center  of  (II)  lie  if 

(a)   G  =  0?  (c)7=0?  (e)  H  =  I=0? 

{h)  H  =  0?  {d)  G=R=0?  {t)  I=G  =  0? 

4.  Show  that  a  sphere  is  determined  by  four  conditions  and  formulate  a 
rule  by  which  to  find  its  equation. 

5.  Find  the  equation  of  the  sphere  which 

(a)  has  the  center  (3,  0,  —  2)  and  passes  through  (1,  6,  —  5). 

Ans.    x2  +  y^  +  z^-Qx-\-4z-3G  =  0. 

(b)  passes  through  the  points  (0,  0,  0),  (0,  2,  0),  (4,  0,  0),  and  (0,  0,  -  6). 

Ans.   x^  -]-y^  +  z^  -4:X-2y  +  6z  =  0. 

(c)  has  its  center  on  the  F-axis  and  passes  through  the  points  (0,  2,  2)  and 
(4,  0,  0).  Ans.   x2  -f  y2  _|.  22  _[.  4  y  _  16  =  0. 

(d)  passes  through  the  points  (1,  1,  0),  (0, 1, 1),  and  (1,  0,  1)  and  whose 
radius  is  11.  Ans.   x2  4-  y2  _}_  ^2  _  14  x  -  14  2/  -  14  z  +  26  =  0. 

(e)  has  the  line  joining  (4,  -  6,  5)  and  (2,  0,  2)  as  a  diameter. 

Ans.   x2  +  ?/2-^22_6x  +  62/-7z-Hl8  =  a 


382  ANALYTIC  GEOMETRY 

6.  Given  two  spheres  Si  :  x"^  +  ij^  +  z"-  -^  GiX  +  Hiy  +  Iiz  +  -^i  =  0  and 
>S2  :  x2  +  ?/2  -f  2;2  +  G2X  +  HiV  +  hz  +  7^2  =  0 ;  show  that  the  locus  of 

Sk  :  x2  +  2/2  +  z2  +  Gix  +  Ihy  +  hz  +  K^ 

+  /b  (x2  +  2/2  +  2;2  +  G2X  +  H^y  +  J22;  +  X2)  =  0 
is  a  circle  except  when  k  =  —  1.     In  this  case  the  locus  is  a  plane  called  the 
radical  plane  of  Si  and  S^. 

7.  The  center  of  the  sphere  Sk  in  problem  6  lies  on  the  line  of  centers  of  Si 
and  S-i  and  divides  it  into  segments  whose  ratio  is  equal  to  k. 

8.  The  equation  of  the  radical  plane  of  Si  and  ^2  (problem  6)  is 

((?i  -  G2)x  +  {Hi-H2)y-\-  {Ii-h)z  +  {K1-K2)  =  0. 

9.  The  radical  plane  of  two  spheres  is  perpendicular  to  their  line  of 
centers. 

10,  The  radical  planes  of  three  spheres  taken  by  pairs  intersect  in  a  line 
perpendicular  to  their  plane  of  centers  which  is  called  the  radical  axis  of  the 
spheres, 

11,  The  radical  planes  of  four  spheres  taken  by  pairs  intersect  in  a  point 
Called  the  radical  center  of  the  spheres. 

12,  When  two  spheres  Si  and  S2  (problem  6)  intersect,  the  system  Sk  con- 
sists of  all  spheres  passing  through  their  circle  of  intersection, 

13,  When  the  spheres  Si  and  S2  (problem  6)  are  tangent,  the  system  Sk 
consists  of  all  spheres  tangent  to  Si  and  -82  at  their  point  of  tangency, 

14,  The  equation  of  the  system  Sk  (problem  6)  may  be  written  in  the  form 

x^-\-y^  +  z^  +  k'x  +  K=  0, 
where  ¥  is  an  arbitrary  constant,  if  the  X-axis  is  chosen  as  the  line  of 
centers  and  the  YZ-plane  as  the  radical  plane  of  ^1  and  ^2- 

15,  The  spheres  of  the  system  in  problem  14  have  their  centers  on  the 
X-axis  and 

(a)  pass  through  the  circle  y^  +  z^  +  E  =  0,  x  =  0  ii  K <0. 

(b)-  are  tangent  to  each  other  at  the  origin  if  K  =  0. 

(c)  are  orthogonal  to  the  sphere  x^  -}-  y^  +  z^  =  K  it  K>  0, 

16,  The  product  of  a  secant  of  a  sphere  drawn  from  a  fixed  point  and  its 
external  segment  is  constant. 

17,  Find  the  square  of  the  length  of  a  tangent  from  a  point  Pi  (xi,  2/1,  2:1) 
to  the  sphere  x^  -{-  y^  +  z"^  -\-  Gx  -{-  Hy  -\-  Iz  +  K  =  0. 

Ans.    Xi2  +  2/i2  +  zi^  +  Gxi  +  Hyi  +  Izi  +  K. 

18,  Show  that  the  equations  of  an  inversion  (p.  297)  in  space  are 
x'  y'  ^  z' 

y- 


X'2  +  /2  +  2'2  x'2  +  ?/'2  +  2;'2  x'2  +  ?/'2  +  2'2 


SPECIAL  SURFACES 


383 


19.  Show  that  the  inverse  of  a  plane  is  a  sphere  unless  the  plane  passes 
through  the  origin,  and  that  in  this  case  the  plane  is  invariant. 

20.  Show  that  the  inverse  of  a  sphere  is  a  sphere  unless  it  passes  through 
the  origin,  when  the  inverse  is  a  plane. 

166.  Cylinders. 

Ex.  1.    Determine  the  nature  of  the  locus  ofy^  =  ix. 

Solution.    The  intersection  of  the  surface  with  a  plane  parallel  to  the 
FZ-plane,  x  =  k,  are  the  lines  (Rule,  p.  345) 
(1)     X  =  fc,     y  =  ±2Vk, 
which  are  parallel  to  the  Z-axis 
(Theorem  II,  p.  342) .    If  A;  >  0, 
the  locus  of  equations  (1)  is  a 
pair  of  lines ;  if  A;  =  0,  it  is  a 
single  line  (the  Z-axis) ;   and 
if  Aj  <  0,  equations  (1)  have  no 
locus. 

Similarly,  the  intersection 
with  a  plane  parallel  to  the 
ZX-plane,  y  =  k,  is  a,  straight 
line  whose  equations  are  (Rule, 
p.  345) 

X  =  ^  A;2,     y  =  k, 

and  which  is  therefore  parallel  to  the  Z-axis. 

The  intersection  with  a  plane  parallel  to  the  JTZ-plane  is  the  parabola 

z  =  k,     2/2  =  4  X. 
For  different  values  of  k  these  parabolas  are  equal  and  placed  one  above 
another. 

It  is  therefore  evident  that  the  surface  is  a  cylinder  whose  elements  are 
parallel  to  the  Z-axis  and  intersect  the  parabola  in  the  XF-plane 
2/2  _  4  a;,     z  =  0. 

It  is  evident  from  Ex.  1  that  the  locus  of  any  equation  which 
contains  but  two  of  the  variables  x,  y,  and  z  will  intersect  planes 
parallel  to  two  of  the  coordinate  planes  in  one  or  more  straight 
lines  parallel  to  one  of  the  axes  and  planes  parallel  to  the  third 
coordinate  plane  in  equal  curves.  Such  a  surface  is  evidently  a 
cylinder.     Hence 

Theorem  IV.  The  locus  of  an  equation  in  which  one  variable  is 
lacking  is  a  cylinder  whose  elements  are  parallel  to  the  axis  along 
which  that  variable  is  measured. 


88-i 


ANALYTIC  GEOMETKY 


167.  The  projecting  cylinders  of  a  curve.  The  cylinders  whose 
elements  intersect  a  given  curve  and  are  parallel  to  one  of  the 
coordinate  axes  are  called  the  projecting  cylinders  of  the  curve. 
Their  equations  may  be  found  by  elimiDating  in  turn  each  of  the 
variables  x,  y,  and  z  from  the  equations  of  the  curve;  for  if  we 
eliminate  z,  for  example,  the  result  is  the  equation  of  a  cylinder 
(Theorem  IV)  which  passes  through  the  curve,  since  values  of  x, 
y,  and  z  which  satisfy  each  of  two  equations  satisfy  an  equation 
obtained  from  them  by  eliminating  one  variable. 


The  equations  of  two  of  the  pi:ojecting  cylinders  may  be  con- 
veniently used  as  the  equations  of  the  curve.* 
The  figure  shows  the  curve  whose  equations  are 

2if-  +  z'^-\-^x--=^z,    ?/2  +  322_8a;  =  i2z.  ^ 
Eliminating  x,  y,  and  z  in  turn,  we  obtain  the  equations  of  the  projecting 


cylinders 


4  2, 


4x=  4z, 


+  4X: 


The  figure  shows  the  first  and  third  of  these  cylinders. 

If  the  curve  lies  in  a  plane  parallel  to  one  of  the  coordinate 
planes,  then  two  of  these  cylinders  coincide  with  the  plane  of  the 
curve,  or  part  of  it. 


*  In  general,  the  equations  of  a  curve  may  be  replaced  by  any  two  independent  equa- 
tions to  which  they  are  equivalent,  that  is,  by  two  independent  equations  which  are 
satisfied  by  all  values  of  x,  y,  and  z  satisfying  the  equations  of  the  curve,  and  only  by 
such  values. 


SPECIAL  SURFACES 


385 


The  projecting  cylinders  of  a  straight  line  are  evidently  planes. 
The  equations  of  a  line  in  terms  of  its  projecting  cylinders  or 
planes  have  already  been  given  (Theorem  IV,  p.  367). 

168.  Cones. 

Ex.  1.    Determine  the  nature  of  the  locus  of  the  equation  16  x^  +  y2  _  z^=0. 

Solution.    Let  Pi  {xi,  ?/i,  Zi)  be  a  point  on  a  curve  C  in  which  the  locus 
intersects  any  plane,  for  example  z  =  k.     Then 
(1)  16  xi2  +  yi2  -  zi^  =  0,     zi  =  k. 

The  origin  0  lies  on  the  surface  (Theorem  III,  p.  345).     We  shall- show 
that  the  line  OPi  lies  entirely  on  the  surface. 
The  direction  cosines  of  OPi  are  (Corollaries, 


Xi 


pp.   332    and    331)   — ,    — ,    and  -,    where 

Pi     Pi  Pi 

Pi2  ^  xi"^  +  2/i2  +  zi"^  =  0Pi2.  Hence  the  coordi- 
nates of  any  point  on  OPi  are  (Theorem  V, 
p.  369) 


Xi 
x  =  —  p, 
PI 


Pi 


z  =  -p. 
PI 


Substituting  these  values  of  cc,  y,  and  z  in 
the  given  equation,  we  obtain 


(2) 


16 


a^lV^       yi2^2       2;i2p2 


+ 


0. 


pr         pr         pi'' 

This  is  true  for  all  values  of  p  since  it  may 
be  obtained  from  the  first  of  equations  (1)  by 

multiplying  by  —  •    Hence  every  point  on 

OPi  lies  on  the  surface,  that  is,  the  entire 

line  lies  on  the  surface.     Hence  the  surface 
is  a  cone  whose  vertex  is  the  origin. 

The  essential  thing  in  the  solution 
of  Ex.  1  is  that  (2)  may  be  obtained 
from  the  first  of  equations  (1)  by  multiplying  by  a  power  of  —  • 

This  may  be  done  whenever  the   equation   of   the   surface   is 
homogeneous*  in  the  variables  x,  y,  and  z.     Hence 

Theorem  V.    The  locus  of  an  equation  which  is  homogeneous  in 
the  variables  ic,  y,  and  z  is  a  cone  whose  vertex  is  the  origin. 

*  An  equation  is  homogeneous  in  x,  y,  and  z  when  all  the  terms  in  the  equation  are  of 

the  same  degree  (footnote,  p.  17). 


386 


ANALYTIC  GEOMETRY 


PROBLEMS 

1 .  Determine  the  nature  of  the  following  loci ;  discuss  and  construct  them. 

(a)  a;2  ^  y%  =  36.  (e)  a;2  _  2/2  +  36  22  =  0. 

(b)  x2  +  2/2  =  22.  (f )  2/2  -  16  x2  +  4  z2  =  0. 

(c)  ?/2  +  4 2;2  =  0.  (g)  x2  +  16?/2  -  4x  =  0. 


(d)    X2  -  Z2  3::  16. 


(h)  x2  +  yz 


2.  Find  the  equations  of  the  cylinders  whose  directrices  are  the  following 
curves  and  whose  elements  are  parallel  to  one  of  the  axes. 

(a)  2/2  +  22  _  4  y  =  0,  a;  =  0.  (c)  Wx'^  -  a'^y'^  =  a%^,  z  =  0. 

(b)  z2  _|.  2x  =  8,  2/  =  0.  (d)  2/2  +  2pz  =  0,  x  =  0. 

3.  Find  the  equations  of  the  projecting  cylinders  of  the  following  curves 
and  construct  the  curve  as  the  intersection  of  two  of  these  cylinders. 


(a)  x2  +  y2  _^2;2  =  25,  x2  +  4  2/2  -  z2  =  0. 

(b)  x2  +  42/2  -  22  =  16,  4x2  +  2/2  +  z2  =: 
^c)  x2  +  2/2  =  4  z,  x2  -  2/2  =  8  z. 

(d)  x2  +  2 2/2  +  4 22  =  32,  x2  +  4 2/2  =  4z. 

(e)  2/2  +  zx  =  0,  2/2  +  2  X  +  2/  -  z  =  0. 


16. 


v 


4.  Discuss  the  following  loci. 

(a)  x2  +  2/2  =  22tan2  7. 

(b)  2/2  +  z2  =  ic2tan2a:. 

(c)  22  +  x2  =  2/2tan2^. 


(d)  x2  +  2/2  =  r2. 

(e)  2/2  +  22  rz  r2. 

(f)  z2  +  x2  =  r2. 


169.  Surfaces  of  revolution.  The  surface  generated  by  revolv- 
ing a  curve  about  a  line  lying  in  its  plane  is  called  a  surface  of 
revolution. 

Ex.  1.-  Find  the  equation  of  the  surface  of  revolution  generated  by  revolv- 
ing the  ellipse  x2  +  4  2/2  —  12  x  =  0,  z  =  0  about  the  JT-axis. 

Solution.  Let  P  (x,  2/,  z)  be  any  point  on  the  surface.  Pass  a  plane  through 
P  and  OX  which  cuts  the  surface  along  one  position  of  the  ellipse,  and  in  this 

plane  draw  OF'  perpendicular 

to  OX.    Referred  to  OX  and 

OY'  as  axes,  the  equation  of 

the  ellipse  is  evidently 

x2  +  4  2/'2  -  12  X  =  0. 

But  from  the  right  triangle 
PAB  we  get  y'^  =  y^  +  z''. 

Substituting  in  (l),  we  get 
(2)  x2  +  42/2  +  4z2-12x=:0. 


SPECIAL  SURFACES  387 

This  equation  expresses  the  relation  which  any  point  on  the  surface  must 
satisfy,  and  it  is  easily  shown  that  any  point  whose  coordinates  satisfy  equa- 
tion (2)  lies  on  the  surface.     It  is  therefore  the  equation  of  the  surface. 

The  method  of  the  solution  enables  us  to  state  the 

Rule  to  find  the  equation  of  the  surface  generated  by  revolving 
a  curve  in  one  of  the  coordinate  planes  about  one  of  the  axes  in 
that  plane. 

Substitute  in  thejequation  of  the  curve  the  square  root  of  the  sum 
of  the  squares  of  the  two  variables  not  measured  along  the  axis  of 
revolution  for  that  one  of  these  two  variables  which  occurs  in  the 
equation  of  the  curve. 

If  the  intersections  of  a  surface  with  all  planes  parallel  to  one 
of  the  coordinate  planes  are  circles,  then  the  surface  is  evidently  a 
surface  of  revolution  whose  axis  is  the  coordinate  axis  perpen- 
dicular to  the  planes  of  the  circular  sections.  This  enables  us  to 
determine  whether  or  not  a  given  surface  is  gaseurface  of  revolu- 
tion whose  axis  is  one  of  the  coordinate  axes. 

170.  Ruled  surfaces.  A  surface  generated  by  a  moving  straight 
line  is  called  a  ruled  surface.  If  the  equations  of  a  straight  line 
involve  an  arbitrary  constant,  then  the  equations  represent  a  sys- 
tem of  lines  which  form  a  ruled  surface.  If  we  eliminate  the 
parameter  from  the  equations  of  the  line,  the  result  will  be  the 
equation  of  the  ruled  surface. 

For  if  («!,  yi,  Z\)  satisfy  the  given  equations  for  some  vakie  of  the  parameter, 
they  will  satisfy  the  equation  obtained  by  eliminating  the  parameter,  that  is,  the 
coordinates  of  every  point  on  every  line  of  that  system  satisfy  that  equation. 

Cylinders  and  cones  are  the  simplest  ruled  surfaces. 

Ex.  1.  Find  the  equation  of  the  surface  generated  by  the  line  whose 
equations  are  -^ 

x-{-y  =  kz,    x-y  =  ~z. 
Ic 

Solution.    We  may  eliminate  k  from  these  equations  of  the  line  hy  multi- 
plying them.    This  gives 
(1)  x2  -  y2  ^  z^. 

This  is  the  equation  of  a  cone  (Theorem  V,  p.  385)  whose  vertex  is  the  origin. 
As  the  sections  made  hy  the  planes  x  =  k  are  circles,  it  is  a  cone  of  revolution 
whose  axis  is  the  X-axis. 


388 


ANALYTIC  GEOMETRY 


We  may  verify  that  the  given  line  lies  on  the  surface  (1)  for  all  values 
of  A;  as  follows : 

Solving  the  equations  of  the  line  for  x  and  y  in  terms  of  z,  we  get 


2\        k 
Substituting  in  (1),  we  obtain 


^-\i.-->  -K-D^- 


4^         k)  4V        k) 


an  equation  which  is  true  for  all  values  of  k  and  z,  as  is  seen  by  removing 
the  parentheses.  Hence  every  point  on  any  line  of  the  system  lies  on  (1), 
since  its  coordinates  satisfy  (1). 

Ex.  2.    Determine  the  nature  of  the  surface  z^  —  3  zx  +  8  y  =  0. 
Solution.     The  intersection  of  the  surface  with  the  plane  z  =  k  is  the 
straight  line  (Rule,  p.  345) 

k^-Skx-\-8y  =  0,     z  =  k.         ^ 


Hence  the  surface  is  the  ruled  surface  generated  by  this  line  as  k  varies. 
To  construct  the  surface  consider  the  intersections  with  the  planes  x  =  0  and 
jc  =  8  whose  equations  are  respectively 

x  =  0,     8 y  +  z3  =  0  and  X  =  8,     Sy  -24z  +  z^  =  0. 

Joining  the  points  on  these  curves  which  have  the  same  value  of  z  gives 
the  lines  generating  the  surface. 


SPECIAL  SURFACES  389 

PROBLEMS 

1 .  Find  the  equations  of  the  surfaces  of  revolution  generated  by  revolving 
the  following  curves  about  the  axes  indicated,  and  construct  the  figures. 

(a)  ?/2  =  4a;-16,  JT-axis.  Ans.  y^ -{- z^  =  ix  -  16. 

(b)  x2  +  4  y2  =  iG,  r-axis.  Ans.  x^  +  i  y^ -{- z^  =  16. 

(c)  x2  —  4  2,  Z-axis.  Ans.  x"^  -\-y^  =  iz. 

(d)  x2  -  ?/2  _  16^  r-axis.  Ans.  x^  -  y^  +  z^  =  16. 

(e)  x2  -  2/2  =  16,  X-axis.  Ans.  x^  -  y^  -  z^  =  16. 

(f)  2/2  -^  22  _  25,  Z-axis.  Ans.  x^  +  y^  +  z^  -  25. 

(g)  2/2  =  2j9z,  Z-axis.  Ans.  A  paraboloid  of  revolution,  x^  +  2/2  =  2pz. 

X2         ?/2  X^         ?/2         z2 

(h) 1 —  =1,  JT-axis.    Ans.  An  ellipsoid  of  revolution, h  —  -\ —  =  1. 

a2      62  (j2      52      52 

a;2      1/2  — 

(i)  -^  -  ^  =  1,  r-axis. 

^       ^  a;2      y^      z2 

-4  ns.    An  hyperboloid  of  revolution  of  one  sheet,  — 1 =  1. 

X^       y^  ^'       ^'       «^ 

^  °  iC2         ^2        22* 

^ns.    An  hyperboloid  of  revolution  of  two  sheets, =1. 

a2      52      52 

2.  Show  that  the  following  loci  are  either  surfaces  of  revolution  or  ruled 
surfaces  whose  generators  are  parallel  to  one  of  the  coordinate  planes.  Con- 
struct and  discuss  the  loci. 

(a)  2/2  +  z2  =  4  X.  (e)  4  x2  +  4  2/2  -  2;2  =  16. 

(b)  0:2  -  4  2/2  +  2:2  =  0.  (f )  r,fiy  _  22  =  0. 

(c)  z^-zx  +  y  =  0.  (g)  x2  -h  z2  =  4. 

(d)  x-hf  +  xz  =  y.  (h)  {x^  +  z'^)y  =  ia'^{2a  -  y). 

3.  Verify  analytically  that  a  sphere  is  generated  by  revolving  a  circle 
about  a  diameter. 

4.  Show  that  the  systems  of  spheres  in  problem  15,  p.  382,  may  be  gen- 
erated by  revolving  the  systems  of  circles  in  Theorem  VIII,  p.  144,  about 
the  X-axis. 

6.  Find  the  equation  of  the  surface  of  revolution  generated  by  revolving 
the  circle  ic2  +  2/2  —  2  ax  +  a2  -  r2  =  0  about  the  T-axis.  Discuss  the  sur- 
face when  a>r^  ix  =  r^  and  a<r. 

Ans.  (x2  +  2/2  -j-  2;2  +  a2  _  ,.2)2  =  4  ^2  (x2  +  z2).  When  a  >  r  the  surface 
is  called  an  anchor  ring  or  torus. 

6.  Find  the  equations  of  the  ruled  surfaces  whose  generators  are  the 
following  systems  of  lines,  and  discuss  the  surfaces. 

(a)  x  +  y  =  k,k{x-y)  =  a'^.  Ans.   x'^  -  y"^  -  a2. 

(b)  4x-2  2/  =  A:z,  A:(4x-f-2  2/)  =  z.  Ans.    16x2-4  2/2  =  ^2. 

(c)  x-22/  =  4^2,  A;(x-22/)  =  4.  Ans.   x^-4y^  =  lQz. 

(d)  X  -1-  A;2/  -I-  4  2  =  4  A;,  fcx  -  2/  -  4  fcz  =  4.  Ans.   x^  +  y^  -  16  z^  =  16. 


390  ANALYTIC  GEOMETRY 

7.  Find  the  equation  of  the  cone  whose  vertex  is  the  origin  and  whose 
elements  cut  the  circle  x'-^  +  y'^  =  W,  z  =  2.  Ans.    x^  +  ^/^  —  4  2^  =  0. 

8.  Find  the  equations  of  the  cones  of  revolution  whose  axes  are  the 
coordinate  axes  and  whose  elements  make  an  angle  of  0  with  the  axis  of 
revolution.    Ans.  y^  +  z'^  =  x^  tan2  0 ;  z^  -\-  x^  =  y^  tan2  0 ;  x"^  +  y^  =  z^  tan2  0. 

9.  Find  the  equations  of  the  cylinders  of  revolution  whose  axes  are  the 
coordinate  axes  and  whose  radii  equal  r. 

Ans.    y^ -{- z"^  =  r^  ;  z^  +  x'^  =  r'^;  x^  +  y^  =  r^. 


IP 


CHAPTER   XXI 


TRANSFORMATION  OF  COORDINATES.     DIFFERENT 
SYSTEMS  OF  COORDINATES 

171.  Translation  of  the  axes. 

Theorem  I.  The  equations  for  translating  the  axes  to  a  new 
origin  0' (h,  k,  I)  are 

(I)  x  =  x'-\-7i,     y  =  yf+k,     z  =  z*+l. 

Proof.  Let  the  coordinates 
of  any  point  before  and  after 
the  translation  of  the  axes  be 
(cc,  y,  z)  and  {x\  y\  z')  respec- 
tively. Projecting  OP  and 
OO'P  on  each  of  the  axes 
(Theorem  II,  p.  328),  we  get 
equations  (I).  q.e.d. 

172.  Rotation  of  the  axes. 

Theorem  II.  If  a^,  fi^,  y^  a^,  ft,  ys,  a'^id  a^,  ft,  yg  are  respec- 
tively the  direction  angles  of  three  mutually  ijerpendicular  lines 
0X\  OY',  and  OZ',  then  the  equations  for  rotating  the  axes  to  the 
position  0-X'Y'Z'  are 

(ijc  =  x^  cos  tti  +  2/'  cos  ttg  +  z^  cos  as, 

(II)  J  1/  =  cc'  cos  ^1  +  y^  cos  ft  +  2!'  cos  ft, 

[  s  =  a?'  cos  yi  +  y^  cos  y^  +  z^  cos  yg.^* 

*  By  Theorem  III,  p.  330,  and  Theorem  VI,  p.  335,  we  see  that  the  direction  cosines  of 
OX',  O  Y',  and  OZ'  satisfy  the  six  equations 

cos2  a,  +  cos2  j3i  +  cos2  Yi  =  1,       cos  Oj  cos  Oj  +  cos  /3i  cos  ^^  +  cos  y,  cos  72  =  0. 
cos2  02  +  C082  (8j  +  cos2  V2  =  1  >       ^^s  o^  COS  flg  +  COS  182  COS  ^3  +  cos  72  «os  Vs  =  0, 

C082  flg  +  COS2  Pg  4-  COS2  73  =  1,  cos  Og  COS  O^  +  COS  /Bg  COS  Pi  +  COS  73  COS  Yj  =  0. 

Hence  only  three  of  the  nine  constants  in  (II)  are  independent. 

391 


392 


ANALYTIC  GEOMETRY 


Proof.    Let  the  coordinates  of  any  point  P  before  and  after 

the  rotation  of  the  axes  be  respec- 
tively {x,  y,  z)  and  (x\  y\  z').  Pro- 
jecting OP  and  OA'B'P  on  each  of 
the  axes  OX,  0  Y,  and  OZ,  we  get,  by 
Corollary  I,  p.  328,  and  Theorems  I 
and  II,  p.  328,  equations  (II).     q.e.d. 

Theorem  III.  The  degree  of  an  equa- 
tion is  unchanged  hy  a  transforma- 
tion of  coordinates. 

Hint.  Show  that  any  transformation  of  coordinates  may  be  effected  by  applying 
Theorems  I  and  II  successively,  then  that  the  degree  cannot  be  raised  by  changing  to  new 
coordinates,  and  finally  that  it  cannot  be  lowered. 


Z' 

^ 

\ 

P 

) 

/ 

/ 

y  A    . 

/      ^^ 

B 


PROBLEMS 


V 


1 .  Transform  the  equation  x'^-\-y^  —  ^x-\-2y  — Az-\-l=0  by  trans- 
lating the  origin  to  the  point  (2,  —  1,  —  1).  Ans.   x'^  +  y"^  —  ^z  =  0. 

2.  Transform  the  equation  bx^  ^Sy"^  +  5z^ —  4:yz +  Szx-\-4:Xy  —  ix  +  2y 
-f  4  2  =  0  by  rotating  the  axes  to  a  position  in  which  their  direction  cosines 
are  respectively  f ,  f^  i .  i,  _  2^  i ;  2^ _  i,  _  i.        Ans.   Sx^  +  Sy^  =  2 z. 

3.  Formulate  a  rule  by  which  to  simplify  a  given  equation  (a)  by  trans- 
lating the  axes,  (b)  by  rotating  the  axes.  How  many  terms  may,  in  general, 
be  removed  from  a  given  equation  by  a  general  transformation  of  coordinates  ? 

4.  Derive  the  equations  for  rotating  the  axes  through  an  angle  6  about 
(a)  the  Z-axis,  (b)  the  X-axis,  (c)  the  F-axis.  rx  =  x'  cosd  —  y' sin d, 

Ans.    (a)  <  y  =  x'  sin  Q  -\-  y'  cos  ^, 
Vz  —  z'. 

5.  Simplify  the  following  equations  by  translating  the  axes  or  by  rotating 
them  about  one  of  the  coordinate  axes. 

(a)  x2  +  2/^  -  2;2  _  6  x  -  8  ^  -h  10  z  =  0.  Ans. 

(b)  3x2- 8x?/ -32/2  _  5^2  +  5  :=0.  Ans. 

(c)  ?/2  +  4  2;2  _  16 x  -  6  ?/  +  16  z  -}-  9  =  0.  Ans. 

(d)  2x2  -  5?/2  -  5^2  -  62/z  =  0.  Ans. 

(e)  9x2-25y2_i_i6z2_242x-80x-60z  =  0.  Ans.   x2  -  2/2  =  4 z. 

6.  Show  that  Ax  -\-  By  -\-  Cz  +  D  =z  0  may  be  reduced  to  the  form  x  =  0 
by  a  transformation  of  coordinates. 

Hint.  Remove  the  constant  term  by  translating  the  axes,  then  remove  the  z-term 
by  rotating  the  axes  about  the  T-axis,  and  finally  remove  the  y-term  by  rotating  about 
the  ^-axis. 


y 


22  =  0. 


x^  -y^  +  z^  =  1. 

?/2  +  4z2=:16x. 
X2  _  4  y2  -  2;2  =r  0. 


TRANSFORMATION  OF  COORDINATES 


393 


7.  Show  that  the  xy-term  may  always  be  removed  from  the  equation 
Ax^  +  By^  +  Cz^  +  Fxy  -\-K  =  0  by  a  rotation  of  the  axes. 

8.  Show  that  the  y2;-term  may  always  be  removed  from  the  equation 
Ax^  +  By^  +  Cz'^  +  Dyz  +  K  =  0  by  rotating  the  axes, 

9.  What  are  the  direction  cosines  of  OX,  OZ,  and  OZ  (Fig.,  p.  392) 
referred  to  0X%  0Y\  and  OZ'?    What  six  equations  do  they  satisfy? 

10.  Show  that  the  six  equations  obtained  in  problem  9  are  equivalent  to 
the  six  equations  in  the  footnote,  p.  391. 

11.  If  (x,  y,  z)  and  {x%  y',  z')  are  respectively  the  coordinates  of  a  point 
before  and  after  a  rotation  of  the  axes,  show  that 

x2  +  y2  +  2;2  ^  X'2  -f-  2/^2  +  2/2, 

173.  Polar  coordinates.  The  line  OP  drawn  from  the  origin 
to  any  point  P  is  called  the  radius  vector  of  P.  Any  point  P 
determines  four  numbers,  its  radius  vector  p  and  the  direction 
angles  of  OP,  namely,  a,  yS,  and  y,  which  are  called  the  polar 
coordinates  of  P. 

These  numbers  are  not  all  independent 
since  a,  jS,  and  7  satisfy  (III),  p.  330.  If  two 
are  known,  the  third  may  then  he  found,  hut 
all  three  are  retained  for  the  sake  of  symmetry. 

Conversely,  any  set  of  values  of 
p,  a,  /?,  and  y  which  satisfy  (III), 
p.  330,  determine  a  point  whose  polar 
coordinates  are  p,  or,  /8,  and  y.  / 

Projecting  OP  on  each  of  the  axes, 
we  get,  by  Corollary  I,  p.  328,  and  Theorem  I,  p.  328, 

Theorem  IV.  The  equations  of  transformation  from  rectangular 
to  polar  coordinates  are 

(IV)  Qc  =  pcosa,     y  =  pcosfi,     z  =  p  cos  y. 

From  Theorem  (IV),  p.  331,  we  obtain 
(1)  p'  =  Gc'  +  y^  +  z\ 

which  expresses  the  radius  vector  in  terms  of  x,  y,  and  ». 


394 


ANALYTIC  GEOMETRY 


174.  Spherical  coordinates.  Any  point  P  determines  three 
numbers,  namely,  its  radius  vector  p,  the  angle  6  between  the 
radius  vector  and  the  Z-axis,  and  the  angle  <^  between  the  pro- 
jection of  its  radius  vector  on  the 
JTF-plane  and  the  Z-axis.  These  num- 
bers are  called  the  spherical  coordinates 
of  P.  6  is  called  the  colatitude  and  <^ 
the  longitude. 

^       Conversely,  given  values  of  p,  6,  and  <^ 

-^  determine    a    point    P   whose   spherical 
coordinates  are  (p,  0,  cf>). 
Projecting  OP  on  OA,  we  get 

OM  =  p  sin  Oj 

and  then  projecting  OP  and  OMP  on  each  of  the  axes,  we  obtain 

Theorem  V.    The  equations  of  transformation  from  rectangular 
to  spherical  coordinates  are 

(V)        a;  =  />  sin  ^  cos  ^,     y  =  p  sin  ^  sin  ^,     «  =  /)  cos  9. 

The  equations  of  transformation  from  spherical  to  rectangular 
coordinates  may  be  obtained  by  solving  (V)  for  p,  6,  and  <^. 


Z< 


P 


175.  Cylindrical  coordinates.  Any  point  P  (x,  y,  z)  determines 
three  numbers,  its  distance  z  from  the 
XF-plane  and  the  polar  coordinates  (r,  <^) 
of  its  projection  (x,  y,  0)  on  the  ZF-plane. 
These  three  numbers  are  called  the  cylin- 
drical coordinates  of  P.  Conversely,  three 
values  of  r,  <^,  and  z  determine  a  point 
whose  cylindrical  coordinates  are  (r,  <f>,  z). 
From  Theorem  I,  p.  155,  we  have  at  once   ^^ 

Theorem  VI.    The  equations  of  transformation  from  rectangular 
to  cylindrical  coordinates  are 

(VI)  Qc  =  r  cos  <f>,    -y  =  r  sin  ^,     z  =  z. 

The  equations  of  transformation  from  cylindrical  to  rectangu- 
lar coordinates  may  be  obtained  by  solving  (VI)  for  r,  <^,  and  z. 


TRANSFORMATION  OF  COORDINATES  395 

PROBLEMS 

1.  What  is  meant  by  the  "locus  of  an  equation  "  in  the  polar  coordinates 
P,  a,  ^,  and  7  ?  in  tlie  spherical  coordinates  p,  6,  and  0  ?  in  the  cylindrical 
coordinates  r,  0,  and  z  ? 

2.  Show  that  the  locus  of  an  equation  in  polar  coordinates  is  symmet- 
rical with  respect  to  the  pole  if  only  the  form  of  the  equation  is  changed 
when  p  is  replaced  by  —  p ;  with  respect  to  one  of*  the  coordinate  planes 
if  only  the  form  of  the  equation  is  changed  when  a  is  replaced  by  ;r  —  or, 
/3  by  ;r  —  /8,  or  7  by  TT  —  7.  Under  what  conditions  will  it  be  symmetrical 
with  respect  to  each  of  the  rectangular  axes  ? 

3.  Find  rules  by  which  to  determine  when  the  locus  of  an  equation  in 
spherical  or  cylindrical  coordinates  is  symmetrical  with  respect  to  the  origin, 
each  of  the  rectangular  axes,  and  each  of  the  coordinate  planes. 

4.  How  may  the  intercepts  of  a  surface  on  the  rectangular  axes  be  found 
if  its  equation  in  polar  coordinates  is  given?  if  its  equation  in  spherical 
coordinates  is  given  ?  if  its  equation  in  cylindrical  coordinates  is  given  ? 

5.  Transform  the  following  equations  into  polar  coordinates. 

(a)  x2  +  2/2  +  22  ^  25.  Ans.   p  =  5. 

(b)  x2  +  y2  _  ^2  =  0.  Ans.    7  =  -• 

4 

(c)  2x2-2/2-22:^0.  Ans.    a  =  cos-iiV3. 

6.  Transform  the  following  equations  into  spherical  coordinates, 
(a)  x2  +  y^  +  z'^  =  16.  Ans.   p  =  4. 

(b)2x  +  3?/  =  0.  Ans.    0  =  tan-i(-|). 

(c)  3 x2  +  3 2/2  =:  7 2;2.  Ans.    6  =  tan-i  i  V2T. 

7.  Transform  the  following  equations  into  cylindrical  coordinates. 

(a)  5x  — ?/  =  0.  An^.    0=rtan-^5. 

(b)  x2  +  2/2  =  4.  Ans.   r  =  2. 

8.  Find  the  equation  in  polar  coordinates  of 

(a)  a  sphere  whose  center  is  the  pole. 

(b)  a  cone  of  revolution  whose  axis  is  one  of  the  coordinate  axes. 

Ans.   (a)  p  =  constant ;  (b)  a  =  constant,  /3  =  constant,  or  7  =  constant. 

9.  Find  the  equation  in  spherical  coordinates  of 

(a)  a  sphere  whose  center  is  the  origin. 

(b)  a  plane  through  the  Z-axis. 

(c)  a  cone  of  revolution  whose  axis  is  the  Z-axis. 

Ans.    (a)  p  =  constant ;  (b)  0  =  constant ;  (c)  6  =  constant. 


396  ANALYTIC  GEOMETRY 

10.  Find  the  equation  in  cylindrical  coordinates  of 

(a)  a  plane  parallel  to  the  XY-plane. 

(b)  a  plane  through  the  Z-axis. 

(c)  a  cylinder  of  revolution  whose  axis  is  the  Z-axis. 

Ans.  (a)  z  =  constant;  (b)  0  =  constant;  (c)  r  =  constant. 

11.  In  rectangular  coordinates  a  point  is  determined  as  the  intersection 
of  three  mutually  perpendicular  planes  (p.  326).     Show  that 

(a)  in  polar  coordinates  a  point  is  regarded  as  the  intersection  of  a  sphere 
and  three  cones  of  revolution  which  have  an  element  in  common. 

(b)  in  spherical  coordinates  a  point  is  regarded  as  the  intersection  of  a 
sphere,  a  plane,  and  a  cone  of  revolution  which  are  mutually  orthogonal. 

(c)  in  cylindrical  coordinates  a  point  is  regarded  as  the  intersection  of 
two  planes  and  a  cylinder  of  revolution  which  are  mutually  orthogonal. 

12.  Show  that  the  square  of  the  distance  between  two  points  whose  polar 
coordinates  are  (pi,  cri,  j8i,  71)  and  {p2,  a^^  ^21  72)  is 

r2  =  p-^  +  p2^  _  2  P1P2  (cos  ai  cos  a^  +  cos  /3i  cos  ^2.  +  cos  71  cos  72). 

13.  Find  the  general  equation  of  a  plane  in  polar  coordinates. 

Ans.   p  {A  cos  a  -\-  B  cos  jS  +  C  cos  7)  +  D  =  0. 

14.  Find  the  general  equation  of  a  sphere  in  polar  coordinates. 

Ans.   /)2  +  yo  (G^  cos  a  +  JBTcos  /3  +  I  cos  7)  +  ii  =  0. 


CHAPTER  XXII 

QUADRIC  SURFACES  AND  EQUATIONS  OF  THE   SECOND 
DEGREE  IN  THREE  VARIABLES 

176.  Quadric  surfaces.  The  locus  of  an  equation  of  the  second 
degree,  of  which  the  most  general  form  is 

(1)   Ax'^-^Bi/^Cz'^-^Dyz+Ezx+Fxy-{-Gx-\-Hy+rz4K  =  0, 

is  called  a  quadric  surface  or  conicoid. 

Theorem  I.  The  intersection  of  a  quadric  with  any  plane  is  a 
conic  or  a  degenerate  conic. 

Proof.  By  a  transformation  of  coordinates  any  plane  may  be 
taken  as  the  ZF-plane,  z  =  0,  and  referred  to  any  axes  the  equa- 
tion of  a  quadric  has  the  form  (1)  (Theorem  III,  p.  392).  Then 
the  equation  of  the  curve  of  intersection  referred  to  axes  in  its 
plane  is  (Kule,  p.  345) 

Ax^  +  Fxy  4-  By^  -\-  Gx -{- Hy  -j- K  =  0, 

and  the  locus  is  therefore  a  conic  or  a  degenerate  conic  (Theo- 
rem XIII,   p.  196).  Q.E.D. 

Corollary.  The  intersection  of  a  cone  of  revolution  with  a  plane 
is  an  ellipse,  hyperbola,  or  parabola  according  as  the  plane  cuts  all 
of  the  elements,  is  parallel  to  two  elements  (cutting  some  on  one  side 
of  the  vertex  and  some  on  the  other),  or  is  parallel  to  one  element 
(cutting  all  the  others  on  the  same  side  of  the  vertex). 

Theorem  II.  The  intersections  of  a  quadric  with  a  system  of  par- 
allel planes  are,  in  general,  similar  conies. 

Proof.  By  a  transformation  of  coordinates  one  of  the  planes 
of  the  system  may  be  taken  as  the  ZF-plane,  and  hence  the  equa- 
tion of  the  system  is  z  —  k^  while  that  of  the  quadric  has  the 

397 


398  ANALYTIC  GEOMETRY 

form  (1)  (Theorem  III,  p.  392).     Hence  the  equation  of  the  curve 
in  which  the  plane  z  =  k  intersects  the  quadric  is  (Kule,  p.  345) 

(2)  Ax^-\-Fxy+By''  +  {Ek+G)x+{Dk+H)y  +  Ck'^+Ik+K=(). 

For  different  values  of  k  this  equation  represents  a  system  of 
similar  conies^  (Corollary  I,  p.  295).  q.e.d. 

177.  Simplification  of  the  general  equation  of  the  second  degree 
in  three  variables.  If  equation  (1)  be  transformed  by  rotating 
the  axes  (Theorem  II,  p.  391),  it  can  be  shown  that  the  new  axes 
may  be  chosen  so  that  the  terms  in  yz,  zx,  and  xy  drop  out  and 
hence  (1)  reduces  to  the  form 

A  '^2  -f  Bhf  4-  C'^2  _^  Qt^  j^  jj^y  _|_  p^  ^  j^^  ^  Q^ 

Transforming  this  equation  by  translating  the  axes  (Theorem  I, 
p.  391),  it  can  be  shown  that  the  axes  may  be  chosen  so  that  the 
transformed  equation  has  either  the  forni 

(1)  A  V  +  B^'if  +  C"^2  _p  j^v  ^  Q 
or  the  forin 

(2)  A"x^  +  B'Y -\- I"z  =  0. 

If  all  of  the  coefficients  in  (l)'and  (2)  are  different  from  zero, 
(1)  and  (2)  may,  with  a  change  in  notation,  be  respectively  written 
in  the  forms 

(3)  ±$±$±$  =  1, 

(4)  S^S=2- 

*  If  the  invariants  of  (2)  (Theorem  YIII,  p.  275)  for  two  different  values  of  k  are  respec- 
tively A,  H,  ©  and  A',  H',  ©'  then  in  order  that  the  conies  be  similar  the  value  of  A  given 

by  =  —  —  must  be  a  real  number. 

©'       A2  © 

All  the  sections  will  belong  to  the  same  type  because  A  will  have  the  same  sign  for  all 
valvies  of  k.  If  the  sections  are  ellipses,  H  and  ©  have  opposite  signs  (Theorem  IX,  p.  277) 
and  A  will  be  real.  The  same  is  true  if  the  sections  arcparabolas  (p.  279).  If  the  sections 
are  hyperbolas,  then,  in  general,  for  values  of  A;  between  certain  limits  the  hyperbolas  will 
be  similar,  and  for  the  remaining  values  of  k,  exclusive  of  the  limits,  the  sections  will  also 
be  similar  (compare  problem  3,  p.  296). 


QUADRIC  SURFACES  399 

The  purpose  of  the  following  sections  is  to  discuss  the  loci  of 
these  equations,*  which  are  called  central  and  non-central  quadrics 
respectively. 

If  one  or  more  of  the  coefficients  in  (1)  or  (2)  are  zero,  the  locus 
is  called  a  degenerate  quadric. 

If  K"  =  0,  the  locus  of  (1)  is  a  cone  (Theorem  V,  p.  385)  unless  the  signs  of  A'% 
B",  and  C"  are  the  same,  in  which  case  the  locus  is  a  point,  namely,  the  origin. 

If  one  of  the  coefficients  A",  B",  and  C"  is  zero,  the  locus  is  a  cylinder  (Theo- 
rem IV,  p.  ;^3)  whose  elements  are  parallel  to  one  of  the  axes  and  whose  directrix 
is  a  conic  of  the  elliptic  or  hyperbolic  type  (p.  195).  If  K"  =  0,  the  locus  will  be  a 
pair  of  intersecting  planes  or  a  line. 

If  tioo  of  the  coefficients  J.'',  B'\  and  C"  are  zero,  the  locus  is  a,  pair  of  parallel 
planes  (coincident  if  K'^  =  0)  or  there  is  no  locus. 

If  one  of  the  coefficients  in  (2)  is  zero,  the  locus  is  a  cylinder  (Theorem  IV,  p.  383) 
whose  directrix  is  a  parabola  or  a  degenerate  central  conic. 

If  two  of  the  coefficients  are  zero,  the  locus  is  a  pair  of  coincident  planes. 
{A"  and  B'^  cannot  be  zero  simultaneously,  as  the  equation  would  cease  to  be  of 
the  second  degree.) 

PROBLEMS 

1.  Construct  and  discuss  the  loci  of  the  following  equations. 

(a)  9x2  _  362/2  +  4^2  ^  q.  (e)  4y2  _  25  =  0. 

(b)  16x2  _  4^/2  -  z2  =  0.  (f)  3y2  +  7 22  =  0. 

(c)  4x2  +  22  -  10  =  0.  (g)  8 ^2  ^.  25 z  =  0. 

(d)  2/2  -  9  z2  +  36  zz  0.  (h)  z^-\-lQ  =  0. 

3.2  y2  ^2 

2.  Discuss  the  locus  of  the  equation  ±  —  ±  —  ±  —  =  0   (a)  if  all  the 

a2      52      c2 

signs  agree ;  (b)  if  two  signs  are  positive.     When  will  the  locus  be  a  cone  of 
revolution  about  the  X-axis  ?   the  F-axis  ?  the  Z-axis  ? 

3.  Show  geometrically  by  means  of  Theorem  I  that  the  sections  of  a 
cylinder  whose  equation  is  of  the  second  degree  made  by  planes  cutting  all 
of  the  elements  are  conies  of  the  same  type.  Show  also  that  the  orthogonal 
projection  on  a  plane  of  an  ellipse  is  an  ellipse  ;  of  an  hyperbola  is  an  hyper- 
bola ;  and  of  a  parabola  is  a  parabola. 

4.  Show  how  to  find  the  equations  of  the  projections  of  a  curve  upon  the 
coordinate  planes  by  means  of  their  projecting  cylinders. 

5.  Prove  the  Corollary  to  Theorem  I  by  determining  the  nature  of  the 
intersection  of  the  cone  x2  +  y2  _  tan2  7  •  z"^  with  the  plane  x  =  tan  /3  •  z  +  &• 

6.  Prove  the  Corollary  to  Theorem  I  by  transforming  x"^  +  y^  =  tan2  7  .  z^ 
by  rotating  the  axes  about  OY  through  an  angle  0  and  considering  the  sec- 
tions formed  by  the  plane  z'  =  fc  if  ^  =  7- 

*  There  is  a  locus  unless  all  of  the  coefficients  of  (3)  are  negative,  when  there  is  no  locus. 


400 


ANALYTIC  GEOMETRY 


178.  The  ellipsoid  _  +  f-  +  _  =  i.   If  all  of  the  coefficients  in 
a?      W      c^ 

(3),  p.  398,  are  positive,  the  locus  is  called  an  ellipsoid.     A  discus- 
sion of  its  equation  gives  us  the  following  properties. 

1.  The  ellipsoid  is  symmetrical  with  respect  to  each  of  the 
coordinate  planes  and  axes  and  the  origin  (Theorem  IV,  p.  346). 
These  planes  of  symmetry  are  called  the  principal  planes  of  the 
ellipsoid. 

2.  Its  intercepts  on  the  axes  are  respectively  (Eule,  p.  346) 

X  —±.  a,     y  =±b,     z  =:k  c. 

The  lines  AA'  =  2  a,  BB'  =2b,CC'  =  2G  are  called  the  axes  of 
the  ellipsoid. 

3.  Its  traces  on  the  principal  planes  are  the  ellipses  ABA'B', 
BCB'C,  and  ACA'C,  whose  equations  are  (p.  346>' 


x^      z' 


=  \. 


4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 
ZF-plane,  z  —  k,  intersects  the  ellipsoid  is  (Rule,  p.  345) 


(1) 


x"^       ?/  _  k^ 

-2  +  ^  =  1-7 


or 


+ 


r 


(C2   -k"") 


IP- 


=  1. 


X 


The  locus  of  this  equa- 
tion is  an  ellipse,  and  for 
different  values  of  k  the 
ellipses  are  similar.    If  k 
increases  from  0  to  c,  or 
decreases  from  0  to  —  c, 
the  plane  recedes  from  the 
J^F-plane,  and  the  axes  of 
the  ellipse  decrease  from 
2  a   and  2  h  respectively 
to  0  when  the  ellipse  degenerates  (p.  195).     \ik>  c  oy  k  <  —  c, 
there  is  no  locus,  and  hence  the  ellipsoid  lies  entirely  between 
the  planes  ;^  =  ±  c. 


/C^ 

Z-i:).^;r^ 

r^^^^^~-— 

\/r  _:>A 

X'-4'W^ 

i/'jo  ; ^ 

^^..^r7 

\  i  ''   _---^ 

^-m 

IP 


QUADRIC  SURFACES 


401 


In  like  manner  the  sections  parallel  to  the  YZ-  and  ZX-planes 
are  similar  ellipses  whose  axes  decrease  as  the  planes  recede,' and 
the  ellipsoid  lies  entirely  between  the  planes  x  =  ±a  and  y  =  ±b. 
Hence  the  ellipsoid  is  a  closed  surface. 

If  a  =  bj  the  section  (1)  is  a  circle  for  values  of  k  such  that 
—  c  <k  <c,  and  hence  the  ellipsoid  is  an  ellipsoid  of  revolution 
whose  axis  is  the  Z-axis.  If  ^  =  c  or  c  =  a,  it  is  an  ellipsoid  of 
revolution  whose  axis  is  the  X-  or  F-axis. 

\i  a  =  b  =  c,  the  ellipsoid  is  a  sphere,  for  its  equation  may  be 
written  in  the  form  x^  +  y"^  -{-  z^  —  a^. 


179.  The  hyperboloid  of  one  sheet  3:2  +  72 


;  =  1.      If  two   of 

a"      o"      c 

the  coefficients  in  (3),  p.  398,  are  positive  and  one  is  negative,  the 

locus  is   called  an  hyperboloid  of  one 

sheet.     Consider  first  the  equation  ^"'^ 


(1) 


r 
b'' 


=  1. 


A  discussion  of  this  equation  gives 
us  the  following  properties. 

1.  The  hyperboloid  is  symmetrical 
with  respect  to  each  of  the  coordinate 
planes  and  axes  and  the  origin  (Theo- 
rem IV,  p.  346). 

2.  Its  intercepts  on  the  X-  and 
F-axes  are  respectively  (Rule,  p.  346) 

a;  =  ±  a,     y  =±b, 

but  it  does  not  meet  the  Z-axis. 

3.  Its  traces  on  the  coordinate  planes  (p.  346)  are  the  conies 


^  _L.  1^'  _  1 
a'-b'~' 


r 

b^ 


^.  =  1. 


of  which  the  first  is  the  ellipse  whose  axes  are  AA'  =2a  and 
BB'  =2b,  and  the  others  are  the  hyperbolas  whose  transverse 
axes  are  BB'  and  A  A'  respectively. 


402  ANALYTIC  GEOMETRY 

4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 
^F-plane,  z  —  h,  intersects  the  hyperboloid  is  (Rule,  p.  345) 

(2)       ^  +  f-:  =  l  +  ^\    or    ^^—^-J^-  =  l. 

-(c2  +  /^0       -(c-'  +  Zc^) 

The  locus  of  this  equation  is  an  ellipse.  If  A;  increases  from 
0  to  00,  or  decreases  from  0  to  —  oo,  the  plane  recedes  from  the 
XF-plane,  and  the  axes  of  the  ellipse  increase  indefinitely  from 
2  a  and  2  h  respectively.  Hence  the  surface  recedes  indefinitely 
from  the  AT-plane  and  from  the  Z-axis. 

In  like  manner  the  sections  formed  by  the  planes  x  =  k^  and 
y  —  k"  are  seen  to  be  hyperbolas.  As  k'  and  k"  increase  numer- 
ically the  axes  of  the  hyperbolas  decrease,  and  when  k'  =  ±  a  oi 
k"  =±b,  the  hyperbolas  degenerate  into  intersecting  lines.  As 
k'  and  k"  increase  beyond  this  point,  the  directions  of  the  trans- 
verse and  conjugate  axes  are  interchanged,  and  the  lengths  of 
these  axes  increase  indefinitely. 

If  either  system  of  hyperbolas  is  projected  orthogonally  on  the  coordinate 
plane  to  which  the  planes  of  the  hyperbolas  are  parallel,  the  projected  system 
will  have  the  appearance  of  the  system  on  p.  201. 

The  hyperboloid  (1)  is  said  to  '^  lie  along  the  Z-axis." 
The  equations 

^   ^  a^       0^       c^   .  a^       ¥       c^ 

are  the  equations  of  hyperboloids  of  one  sheet  which  lie  along 

the  Y-  and  A-axes  respectively. 

If  a  =  &,  the  hyperboloid  (1)  is  a  surface  of  revolution  whose 
axis  is  the  Z-axis,  because  the  section  (2)  becomes  a  circle.  The 
hyperboloids  (3)  will  be  hyperboloids  of  revolution  \i  a  =  c  and 
b  =  c  respectively. 

180.  The  hyperboloid  of  two  sheets  —  —  ^  — 2  =  1*  ^^  ^^^7 
.one  of  the  coefficients  in  (3),  p.  398,  is  positive,  the  locus  is 
called  an  hyperboloid  of  two  sheets.     Consider  first  the  equation 

^  >  a'      6^       c^ 


QUADRIC  SURFACES 


403 


1.  The  hyperboloid  is  symmetrical  with  respect  to  each  of  the 
coordinate  planes  and  axes  and  the  origin  (Theorem  IV,  p.  346). 

2.  Its  intercepts  on  the  X-axis  ai'e  x  =±.a,  but  it  does  not 
cut  the  Y-  and  Z-axes. 

3.  Its  traces  on  the  XY-  and  ZZ-planes  (p.  346)  are  respec- 
tively the  hyperbolas 


r 


1, 


which  have  the  same  transverse  axis  AA'  =  2a,  but  it  does  not 
cut  the  FZ-plane. 

4.  The  equation  of  the  curve  in  which  a  plane  parallel  to 
the  yz-plane,  x  =  k,  intersects  the  hyperboloid  of  one  sheet  is 
(Rule,  p.  345) 


y 


or 


r 


Xk-'-a^) 


1. 


a"  ^  ■        a^^  ^ 

This  equation  has  no  locus  \i  —  a  <  k  <  d.  li  k  =±a,  the 
locus  is  a  degenerate  ellipse,  and  as  k  increases  from  a  to  oo,  or 
decreases  from  —a  to  —  oo,  the 
locus  is  an  ellipse  whose  axes 
increase  indefinitely.  Hence  the 
surface  consists  of  two  branches 
or  sheets  which  recede  indefi- 
nitely from  the  FZ-plane  and 
from  the  X-axis. 

In  like  manner  the  sections,  formed  by  all  planes  parallel 
to  the  XY-  and  ZZ-planes  are  hyperbolas  whose  axes  increase 
indefinitely  as  their  planes  recede  from  the  coordinate  planes. 

The  hyperboloid  (1)  is  said  to  "  lie  along  the  Z-axis." 

The  equations 


(2) 


r 


+ 


are  the  equations  of  hyperboloids  of  two  sheets  which  lie  along 
the  Y-  and  Z-axes  respectively. 


404  ANALYTIC   GEOMETRY 

If  h  =  Cj  c  =  a,  or  a  =  h,  the  hyperboloids  (1)  and  (2)  are 
respectively  hyperboloids  of  revolution. 

It  should  be  noticed  that  the  locus  of  (3),  p.  398,  is  an  ellipsoid  if  all  the  terms 
on  the  left  are  positive,  an  hyperboloid  of  one  sheet  if  but  one  term  is  negative, 
and  an  hyperboloid  of  two  sheets  if  two  terms  are  negative.  If  all  the  terms  on 
the  left  are  negative,  there  is  no  locus.  If  the  locus  is  an  hyperboloid,  it  will  lie 
along  the  axis  corresponding  to  the  term  whose  sign  differs  from  that  of  the  other 
two  terms. 

PROBLEMS 

1.  Discuss  and  construct  the  loci  of  the  following  equations. 

(a)  4x2  +  97/2  +  I6z2  ^  144.  (e)  9x2  _  ^/S  +  9^2  ^  35. 

(b)  4x2  +  9^,2  _  i6;22  ^  144.  (f)  22  _  4x2  _  4^/2  =  \q, 

(c)  4x2  _  9^/2  _  i6z2  :=  144.  (g)  i6a;2  j^yij^iQ^'^^  64. 

(d)  x2  +  16  2/2  +  z2  ^  64.  (h)  x2  +  y2  _  ^2  =  26. 

2.  For  what  values  of  k  or  ¥  will  the  sections  of  the  hyperboloid  of  one 

^2        y2        2;2 

sheet,  — \-~ =  1,  formed  by  the  planes  x  =  k   or  y  —  V  be   similar 

a'      ^"^      ^'  x2     2/2     z2 

hyperbolas  ?    the  hyperboloid  of  two  sheets 1 —  =1? 

a2      62      q1 

3.  Show  analytically  that  the  intersection  of  an  ellipsoid  with  any  plane 
is  a  conic  of  the  elliptic  type. 

4.  Show  analytically  that  the  section  of  an  hyperboloid  of  (a)  one  sheet, 
(b)  two  sheets  formed  by  a  plane  passing  through  the  axis  along  which  the 
hyperboloid  lies,  is  an  hyperbola. 

rffh         2/2         ^2 

5.  Show  that 1 f-  —  =  {Ax  -\-  By  ■\-  Czf  is  the  equation  of  the  cone 

a2      62      Q% 

whose  vertex  is  the  origin  which  passes  through  the  intersection  of  the 

r^  y1  2;2 

ellipsoid f-  —  H —  =  1  and  the  plane  J.x  +  JBy  +  C2;  =  1. 

a2      62      c2 

6.  Show  that  xsf  i  -  i^  +  2/2('i  - -")  +  ^2/1  _  i\  =  0  is  the  equa- 

\a2      r2/  \62      r2/  \c2      r'^)  ^ 

tion   of  the  cone  whose  vertex  is  the   origin  which  passes  through   the 
intersection  of  the  ellipsoid  and  the  sphere  x2  +  2/2  +  22  =  r^. 

7.  If,  in  problem  6,  a>b>c  and  r  =  6,  show  that  the  cone  degenerates 
into  a  pair  of  planes  whose  intersections  with  the  ellipsoid  are  circles.  What 
is  the  nature  of  the  cone  if  r=:a?  ifr  =  c? 

8.  Find  the  equations  of  the  planes  whose  intersections  with  the  ellipsoid 
9  x2  -I-  25  2/2  +  169  z2  =  1  are  circles.  Ans.    4  x  =  ±  12  z  +  ^. 


QUADRIC  SURFACES 


405 


9.  Find  the  equation  of  the  cone  whose  vertex  is  the  origin  which 

2;2         yl         ^2 

through  the  intersection  of  (a)  the  hyperboloid  of  one  sheet 1-  ^ =  1. 

«        o        o  a^      ^'      c2 

2j2        y1        ^2 

(b)  the  hyperboloid  of  two  sheets  -^  —  rr  — ^  =  1  with  the  sphere  x^  +  y^ 

+  2^  =  r2.    For  what  value  of  r  will  the  cone  degenerate  into  a  pair  of  planes 
whose  intersections  with  the  hyperboloid  are  circles  ? 

An..   (a)x^(l-i)  +  2,^(l-I)-.^(i  +  l)  =  0;r=aif«>6. 

10.  Find  the  equations  of  the  two  systems  of  planes  whose  intersections 
with  (a)  an  ellipsoid,  (b)  an  hyperboloid  of  one  sheet,  (c)  an  hyperboloid  of 
two  sheets,  are  circles. 

2  2 

181.  The  elliptic  paraboloid  ^  +  ^  =  2  cs;.   If  the  coefficient 

a        o 

of  y^  in  (4),  p.  398,  is  positive,  the  locus  is  called  an  elliptic 
paraboloid.     A  discussion  of  its 
equation    gives    us    the    following 
properties. 

1.  The  elliptic  paraboloid  is 
symmetrical  with  respect  to  the 
YZr  and  ZZ-planes  and  the  Z-axis 
(Theorem  IV,  p.  346). 

2.  It  passes  through  the  origin 
(Theorem  III,  p.  345)  but  does  not 
intersect  the  axes  elsewhere  (Rule, 
p.  346). 

3.  Its  traces  on  the  coordinate  planes  (p.  346)  are  respectively 
the  conies  „        „  „ 


0, 


2r^      ^ 


2cz, 


of  which  the  first  is  a  degenerate  ellipse  (p.  195)  and  the  others 
are  parabolas. 

4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 
AT-plane,  z  =  k,  cuts  the  paraboloid  is  (Rule,  p.  345) 


or      ir 


^^^''  ''  2^k 


+ 


?r 


x: 


2b'ck 


1. 


"   i- 


<5) 


406  ANALYTIC  GEOMETRY 

The  curve  is  an  ellipse  if  c  and  k  have  the  same  sign,  but  there 
is  no  locus  if  c  and  k  have  opposite  signs.  Hence,  if  c  is  positive, 
the  surface  lies  entirely  above  the  XF-plane.  If  k  increases 
from  0  to  oc,  the  plane  recedes  from  the  ZF-plane  and  the  axes 
of  the  ellipse  increase  indefinitely.  Hence  the  surface  recedes 
indefinitely  from  the  ZF-plane  and  from  the  Z-axis. 

In  like  manner  the  sections  parallel  to  the  YZ-  and  ZX-planes 
are  parabolas  whose  vertices  recede  from  the  ZF-plane  as  their 
planes  recede  from  the  coordinate  planes. 

The  loci  of  the  equations 

(1)  |!  +  S  =  2-.     5  +  S  =  2*^ 

are   elliptic    paraboloids    which   lie   along    the  X-  and    F-axes 
respectively. 

If  a  =  h,  the  first  surface  considered  is  a  paraboloid  of  revolu- 
tion whose  axis  is  the  Z-axis ;  and  ii  b  =  c  and  a  =  c,  the  parab- 
oloids (1)  are  surfaces  of  revolution  whose  axes  are  respectively 
the  X-  and  T-axes. 

An  elliptic  paraboloid  lies  along  the  axis  corresponding  to  the  term  of  the  first 
degree  in  its  equation,  and  in  the  positive  or  negative  direction  of  the  axis 
according  as  that  term  is  positive  or  negative. 

182.  The  hyperbolic  paraboloid  —-^  =  ^cz.  If  the  coeffi- 
cient of  y"^  in  (4),  p.  398,  is  negative,  the  locus  is  called  an  hyperbolic 
paraboloid. 

1.  The  hyperbolic  paraboloid  is  symmetrical  with  respect  to 
the  YZ-  and  ZA-planes  and  the  Z-axis  (Theorem  IV,  p.'  346). 

2.  It  passes  through  the  origin  (Theorem  III,  p.  345)  but  does 
not  cut  the  axes  elsewhere  (Rule,  p.  346). 

3.  Its  traces  on  the  coordinate  planes  (p.  346)  are  respectively 
the  conies 

of  which  the  first  is  a  degenerate  hyperbola  (p.  195)  and  the 
others  are  parabolas. 


QUADRIC  SURFACES 


407 


4.  The  equation  of  the  curve  in  which  a  plane  parallel  to  the 
ZF-plane,  z  =  kj  cuts  the  paraboloid  is  (Kule,  p.  345) 

—„  —  T7:  =  2ch  or 


2a''ck      2b''ck 
If  c  is  positive,  the  transverse  axis 


The  locus  is  an  hyperbola, 
of  the  hyperbola  is  parallel  to  the 
.Y-  or  F-axis  according  as  k  is  posi- 
tive or  negative.  If  k  increases 
from  0  to  00,  or  decreases  from  0 
to  —  00,  the  plane  recedes  from  the 
XF-plane  and  the  axes  of  the 
hyperbolas  increase  indefinitely. 
Hence  the  surface  recedes  indefi- 
nitely from  the  ZF-plane  and  the 
Z-axis.    The  surface  has  approximately  the  shape  of  a  saddle. 

In  like  manner  the  sections  parallel  to  the  other  coordinate 
planes  are  parabolas  whose  vertices  recede  from  the  XF-plane  as 
their  planes  recede  from  the  coordinate  planes. 

The  loci  of  the  equations 


..--.  =  ^by,     f,- 


2  ax 


x"       Z' 

are    hyperbolic    paraboloids    lying    along    the    F-   and   Z-axes 
respectively. 

An  hyperbolic  paraboloid  also  lies  along  the  axis  which  corresponds  to  the 
term  of  the  first  degree  in  its  equation. 


PROBLEMS 


1.    Discuss  and  construct  the  following  loci. 


(a)  ?/2  +  z2^4a;. 

(b)  y2_22^4a;. 


(c)  9z2_4a;2 

(d)  16  x2  +  z2  : 


:  288  y. 
64  y. 


2.  Prove  that  the  parabolas  of  the  systems  obtained  by  cutting  (a)  an 
elliptic  paraboloid,  (b)  an  hyperbolic  paraboloid  by  planes  parallel  to  one 
of  the  coordinate  planes,  are  all  equal. 

3.  Show  analytically  that  any  plane  parallel  to  the  axis  along  which 
(a)  an  elliptic  paraboloid,  (b)  an  hyperbolic  paraboloid  lies,  intersects  the 
surface  in  a  parabola. 


408  ANALYTIC  GEOMETRY 

4.  Show  analytically  that  any  plane  not  parallel  to  the  axis  of  an  elliptic 
paraboloid  intersects  the  surface  in  an  ellipse. 

5.  Show  analytically  that  any  plane  not  parallel  to  the  axis  of  an  hyper- 
bolic paraboloid  intersects  the  surface  in  an  hyperbola. 

6.  Find  the  equation  of  the  cone  whose  vertex  is  the  origin  which  passes 

through  the  intersection  of  the  paraboloid 1-  —  =  2  cz  and  the  sphere 

x2  +  2/2  +  ^2  =  2  rz.  ^^ 


^ns.   x2(^^-c)  +  y2(^^-c)-cz2  =  o. 


7.  By  means  of  problem  6  find  the  equations  of  two  systems  of  planes 
whose  intersections  with  the  paraboloid  are  circles. 

183.  Rectilinear  generators.  The  equation  of  the  hyperboloid 
of  one  sheet  (p.  401)  may  be  written  in  the  form 

^^  W"        C"  h''  ]9 

As  this  equation  is  the  result  of  eliminating  h  from  the  equa- 
tions of  the  system  of  lines 

a      G  I  b  J       a      G      ky         b 

the  hyperboloid  is  a  ruled  surface  (p.  387).  Equation  (1)  is  also 
the  result  of  eliminating  k  from  the  equations  of  the  system  of 
lines  /  \  -,  / 

a      G        ^\         bj       a       c      k\         b 

and  the  hyperboloid  may  therefore  be  regarded  in  two  ways  as  a 
ruled  surface. 

In  like  manner  the  hyperbolic  paraboloid  contains  the  two 
systems  of  lines 

a      b  a      b       k 

T  X   ,   y       .         X       y      2g 

and  -  -\-  ^  —  kz, 7  =  -^^ 

a      b  a       b        k 

These  lines  are  called  the  rectilinear  generators  of  these  surfaces. 
Hence 

Theorem  III.  The  hyperboloid  of  one  sheet  and  the  hyperbolic 
paraboloid  have  two  systems  of  reetilinear  generators,  that  is,  they 
may  be  regarded  in  tivo  ways  as  mled  surfaces. 


T^LATK    IT 


Elliptic  Paraboloid  Hyperbolic  Paraboloid 

Non-Central  Quadrics 


Hyperboloid  of  one  sheet  Hyperbolic  Paraboloid 

Riled  Quadrics 


V 


QUADRIC  SURFACES  409 

MISCELLANEOUS   PROBLEMS 

1.  Construct  the  following  surfaces  and  shade  that  part  of  the  first  inter- 
cepted by  the  second. 

(a)  ^2  +  4  y2  _|.  9  2;2  =  36,  x2  +  y^  +  2;2  =  16. 

(b)  x2  +  2/2  _|.  2;2  ^  64,  x2  +  2/2  -  8x  =  0. 

(c)  4  x2  +  2/2  -  4  z  =  0,  x2  +  4  2/2  -  22  =  0. 

2.  Construct  the  solids  bounded  by  the  surfaces  (a)  x2  +  2/2  =  a2,  z  =  mx, 
2  =  0 ;   (b)  x2  +  2/2  =  az,  x2  +  2/2  =  2  ox,  z  =  0. 

3.  Show  that  two  rectilinear  generators  of  (a)  an  hyperbolic  paraboloid, 
(b)  an  hyperboloid  of  one  sheet,  pass  through  each  point  of  the  surface. 

4.  If  a  plane  passes  through  a  rectilinear  generator  of  a  quadric,  show 
that  it  will  also  pass  through  a  second  generator  and  that  these  generators  do 
not  belong  to  the  same  system. 

5.  The  equation  of  the  hyperboloid  of  one  sheet  (p.  401)  may  be  written 

in  the  form =  1 By  treating  this  equation  as  we  treated  equa- 

62      c2  a2 

tion  (1),  p.  408,  we  obtain  the  equations  of  two  systems  of  lines  on  the  sur-' 
face.  Show  that  these  systems  of  lines  are  identical  with  those  already 
obtained. 

6.  Show  that  a  quadric  may,  in  general,  be  passed  through  any  nine 
points. 

7.  If  a  >  6  >  c,  what  is  the  nature  of  the  locus  of 

x2  2/^         j         2^        ^-, 


a2  -  X      62  _  X      c2  -  \ 
ifX>a2?  ifa2>X>62?  if62>X>c2?  ifX<c2? 

8,  Show  that  the  traces  of  the  system  of  quadrics  in  problem  7  are  confogal 
conies. 

9.  Show  that  every  rectilinear  generator  of  the  hyperbolic  paraboloid 

/J.2        1/2  X        V 

—  =  2  cz  is  parallel  to  one  of  the  planes  -  ±  -  =  0. 

a^      b'2  ah 

10.  Prove  that  the  projections  of  the  rectilinear  generators  of  (a)  the 
hyperboloid  of  one  sheet,  (b)  the  hyperbolic  paraboloid,  on  the  principal 
planes  are  tangent  to  the  traces  of  the  surface  on  those  planes. 

11.  A  plane  passed  through  the  center  and  a  generator  of  an  hyperboloid 
of  one  sheet  intersects  the  surface  in  a  second  generator  which  is  parallel  to 
the  first. 

12.  Show  how  to  generate  each  of  the  central  quadrics  by  moving  an 
ellipse  whose  axes  are  variable. 

13.  Show  how  to  generate  each  of  the  paraboloids  by  moving  a  parabola. 


CHAPTER  XXIII 

RELATIONS  BETWEEN  A  LINE  AND  QUADRIC.     APPLICA- 
TIONS OF  THE  THEORY  OF  QUADRATICS 

184.  The  equation  in  p.  Relative  positions  of  a  line  and  quadric.  Con- 
sider any  equation  of  the  second  degree,  whose  locus  is  a  quadric  surface, 
degenerate  or  non-degenerate,  and  a  line  whose  parametric  equations  are 
(Theorem  V,  p.  369) 

(1)  x  =  Xi-i-  p  cos  a,  y  —  yi  +  p  cos  ^,  z  =  Zi  +  p  cos  y. 

If  these  values  of  x,  y,  and  z  satisfy  the  equation  of  the  quadric,  then  the 
point  P(x,  y,  z)  on  the  line  (1)  will  also  lie  on  the  quadric.  Substituting 
from  (1)  in  the  equation  of  the  quadric  and  arranging  the  result  according 
to  powers  of  p,  the  result  is  a  quadratic 

(2)  Ap^ -\- Bp -\- C  =  0 

whose  roots  are  the  directed  distances  from  Pi  (xi,  2/1,  zi)  to  the  points  of 
intersection  of  the  line  (1)  and  the  quadric.  The  quadratic  (2)  is  called 
the  equation  in  p  for  the  given  quadric  (compare  §  94,  p.  235).  Hence  we 
have  the 

Rule  to  derive  the  equation  in  p  for  any  quadric. 

Substitute  the  values  of  x,  y,  and  z  given  by  (1)  in  the  equation  of  the  quadric 
and  arrange  the  result  according  to  powers  of  p. 

"Denoting  the  discriminant  of  (2),  B^  —  4lAC,  by  A,  it  is  evident  from 
Theorem  II,  p.  3,  that 

(a)  the  line  is  a  secant  of  the  quadric  if  A  is  positive. 

(6)  the  line  is  tangent  to  the  quadric  if  A  is  zero. 

(c)  the  line  does  not  meet  the  quadric  if  A  is  negative. 

If  C  =  0,  one  root  of  (2)  is  zero  (Case  I,  p.  4),  and  hence  P^  lies  on  the  qviadric. 

If  B  =  0,  the  roots  of  (2)  are  numerically  equal  with  opposite  signs  (Case  II,  p.  4)  and  P, 
is  the  middle  point  of  the  chord  formed  by  (1). 

If  ^  =  0,  one  root  of  (2)  is  infinite  (Theorem  IV,  p.  15)  and  the  line  is  said  to  intersect 
the  quadric  at  infinity. 

If  B  =  C  =  0,  both  roots  are  zero  (Case  III,  p.  5)  and  the  line  is  said  to  be  tangent  to  the 
quadric  at  P^. 

If  A  =  B  =  0,  both  roots  are  infinite  and  the  line  is  said  to  be  tangent  to  the  quadric  at 
infinity. 

If  A  =  B  =  C  =  0,  any  number  is  a  root  of  (2),  and  hence  all  points  on  the  line  lie  on  the 
quadric  (compare  p.  226). 

410 


LINE  AND  QUADRIC  411 

PROBLEMS 

1.  Determine  the  relative  positions  of  the  following  lines  and  quadrics. 

(a)  X  =  -  6  +  I/),  2/  =  6  -  f /),  z  =  3  -  ip,  x2  +  2/2  +  4z2  =  le. 

Ans.    Secant. 

(b)  X  =  f  /3,  y  =  9  +  ^  /3,  z  =  1  -  f  /),  2/^  +  4  z2  =  8  X.      Ans.   Do  not  meet. 

(c)  X  =  4  +  f  p,  2/  =  -  2  +  |p,  z  =  5  +  ip,  x2  +  2/2  +  2;2  ^  36. 

Ans.    Tangent. 

(d)  X  =  3  +  i  ^P,  y  =  I  +  i  V3p,  z  =  _  2  -  1  V3p,  x2  -  z2  ^  2 2/. 

J.ns.    Line  lies  on  quadric. 

(e)  ?^  =  ^  =  ^^,  x2  +  4 2/2  -  z2  _  4 a;  =  0.  Ans.   Secant. 

z  o  t) 

(f)  ^^  =  ^  =  ^^,  x2  +  42/2  -  9z2  =  36. 

9         6        —  5 

Ans.    Secant  with  one  point  of  intersection  at  infinity. 

2.  Find   the    condition    that   the   line    x=2+pcosar,    2/  =  l+PCOS/3, 
z  =  —  1  +  p  cos  7  should  be  tangent  to  the  paraboloid  x2  —  2/2  +  3  z  =  0. 

Ans.   4  cos  a:  —  2  cos  /3  —  3  cos  7  =  0. 

3.  Find  the  condition  that  Pi  (xi,  yi,  Zi)  should  be  the  middle  point  of  the 
chord  of  the  hyperboloid  x2  —  2/2  +  4  z2  =  16  formed  by  the  line  x  =  Xi  +  |  p, 

Ans.   2  xi  +  2/1  —  8  zi  =  0. 

185.  Tangent  planes.   Consider  the  elliptic  paraboloid 
x2      v2 
(>>  a^  +  l  =  "'' 

and  the  line 
(2)  X  =  Xi  +  p  cos  a,     2/  =  2/1  +  p  cos  i3,     z  =  Zi  +  p  cos  7. 

Substituting  from  (2)  in  (1),  we  obtain  the  equation  in  p  (p.  410) 

If  Pi(xi,  2/1,  zi)  is  to  lie  on  (1),  and  (2)  is  to  be  tangent  to  (1)  at  Pi,  both 
roots  of  (3)  must  be  zero,  and  hence  (Case  III,  p.  6) 

scicosa      2/1  cos /3  .      Xi*  ,  2/1^      „  ^ 

(4)  .J__  +  ?^-ccos7  =  0,    _  +  _-2«i  =  0. 

Solving  (2)  for  the  direction  cosines,  we  get 

X  -Xi  ^      y  -Vi  z  —  z\ 

(5)  cos  a  — -,     cos  /3  = ,     cos  7  = • 


412  ANALYTIC  GEOMETRY 

Substituting  from  (5)  in  the  first  of  equations  (4),  we  get 

(Q)  ^i(^-'gi)  ,  y{y-  yi)  ^  c  (g  -  ^i) 

as  the  condition  that  P{x,  y,  z)  should  lie  on  a  line  tangent  to  (1)  at  Pi. 
Simplifying  (6)  by  means  of  the  second  of  equations  (4),  we  obtain 

(')  f  +  f^ -(--)• 

This  is  the  equation  of  a  plane  (Theorem  II,  p.  349).  Hence  all  of  the 
lines  tangent  to  (1)  at  Pi  lie  in  a  plane  which  is  called  the  tangent  plane. 

This  method  may  be  summed  up  in  the 

Rule  to  derive  the  equation  of  the  plane  which  is  tangent  to  a  quadric  at  a 
given  point  Pi(xi,  y^  ^i)- 

First  step.  Derive  the  equation  in.p  and  set  the  coefficient  of  p  and  the 
constant  term  equal  to  zero. 

Second  step.  Solve  the  parametric  equations  of  the  line  for  its  direction 
cosines  and  substitute  in  the  first  equation  obtained  in  the  first  step. 

Third  step.  Simplify  the  equation  obtained  in  the  second  step  by  means  of 
the  second  equation  obtained  in  the  first  step.  The  result  is  the  required 
equation. 

By  means  of  this  Rule  we  obtain 

Theorem  I.     The  equation  of  the  plane  which  is  tangent  at  Pi  (xi,  ?/i,  Zi)  to  the 

central  quadric  ±  _  ±  |- ±  _  =  1  is  ±  ^  ±  ^  ± -J^  =  1; 

x2     7/2  sc.ac      y^y 

non-central  quadric       —±  —  =  ^cz      is       — —  4-  — —  =  c{z  -\-  Zj). 
a^     b^  a  b 

Theorem  n.  The  equation  of  the  plane  which  is  tangent  to  any  quadric  at 
Pii^h  2/ii  ^i)  *s  found  by  substituting  X\X,  yiy,  and  ZiZ  for  x^,  y^,  and  z^; 
i  {yix  +  xiy),  i  {ziy  +  yiz),  and  i  {xiz  +  zix)  for  xy,  yz,  and  zx;  and  i  (x  +  Xi), 
i  (y  +  2/i)»  ttw<^  H^  +  ^i)  f^'"'  ^»  y^  ^"•^  g  ^^  ^^^  equation  of  the  quadric. 

186.  Polar  planes.  If  Pi  is  a  point  on  a  quadric,  the  equation  of  the 
tangent  plane  at  Pi  may  be  found  by  Theorem  II.  If  Pi  is  not  on  the  quad- 
ric, the  plane  found  by  Theorem  II  is  called  the  polar  plane  of  Pi,  and  Pi  is 
called  the  pole  of  that  plane. 

In  particular,  the  polar  plane  of  a  point  on  a  quadric  is  the  plane  tangent 
to  the  quadric  at  that  point,  and  the  pole  of  a  tangent  plane  is  the  point  of 
tangency. 

187.  Circumscribed  cones.  All  of  the  lines  passing  through  a  point  not 
on  a  given  quadric  which  are  tangent  to  the  surface  form  a  cone  which  is 
said  to  be  circumscribed  about  the  quadric. 


LINE  AND  QUADIUC 


413 


Ex.  1.    Find  the  equation  of  the  cone  circumscribed  about  the  ellipsoid  a;2  +  3  ^2 
+  3  ^2  =  9  whose  vertex  is  the  point  P^  (4,  —  2,  4) . 

Solution.   The  parametric  equations  of  any  line  through  P^  are  (Theorem  V, 
p.  369) 

(1)  a;  =  4  + /)  cos  a,    y  =  -  2  +  pcosjS,    2;  =  4  +  /)  cos  7. 

Substituting  these  values  of  x,  y,  and  z  in  the  equation  of  the  ellipsoid,  we  obtain 
the  equation  in  p 

(2)  (cos2  a  +  3  cos2/3  +  3  cos2  7)  p2  _^  (g  cos  or  -  12  cos  /3  +  24  cos  7)  /o  +  67  =  0. 
If  (1)  is  tangent  to  the  ellipsoid,  then  [(6),  p.  410] 

(3)  (8  cos  a  -  12  cos  iS  +  24  cos  7)2  -  4  •  67  (cos2  a  +  3  cos2/3  +  3  cos2  7)  =  0. 


Solving  (1)  for  the  direction  cosines,  substituting  in  (3),  and  multiplying  by  p^, 
we  get 
(4)     [8(a;-4)-12(y  +  2)  +  24(z-4)]2-268[(x-4)2  +  3(y  +  2)2  +  3(z-4)2]  =  0 

as  the  condition  that  P  {x,  y,  z)  should  lie  on  a  line  passing  through  P^  which  is 
tangent  to  the  ellipsoid.    Hence  (4)  is  the  equation  of  the  required  cone. 

That  the  locus  of  (4)  is  really  a  cone  whose  vertex  is  Pi  is  easily  seen  by  moving 
the  origin  to  Pi  and  applying  Theorem  V,  p.  385. 

In  constructing  the  figure,  two  divisions  on  each  axis  were  taken  for  the  unit. 

The  reasoning  employed  in  the  solution  of  Ex.  1  justifies  the 

Rule  to  find  the  equation  of  the  cone  whose  vertex  is  Pi  (xi,  yi,  Zi)  which 
circumscribes  a  given  quadric. 

First  step.    Derive  the  equation  in  p  and  set  its  discriminant  equal  to  zero. 

Second  step.  In  the  result  of  the  first  step  substitute  the  values  of  the  direc- 
tion cosines  of  a  line  through  Pi  obtained  from  the  parametric  equations  of  the 
line.     The  result  is  the  required  equation. 


414  ANALYTIC  GEOMETRY 


PROBLEMS 


,     1.  Prove  that  the  plane  of  the  two  rectilinear  generators  which  pass 
through  any  point  on  a  ruled  quadric  is  the  tangent  plane  at  that  point. 

2.  Prove  that  every  plane  which  passes  through  a  rectilinear  generator  of 
a  ruled  quadric  is  tangent  to  the  quadric  at  some  point  of  that  generator. 

3.  Prove  analytically  that  every  plane  tangent  to  a  cone  passes  through 
the  vertex. 

4.  Prove  that  the  polar  plane  of  any  point  in  a  given  plane  passes  through 
the  pole  of  that  plane. 

5.  Prove  that  the  pole  of  any  plane  which  passes  through  a  given  point 
lies  in  the  polar  plane  of  that  point. 

6.  Prove  that  the  curve  of  contact  of  a  cone  circumscribed  ahout  a 
quadric  lies  in  the  polar  plane  of  the  vertex. 

7.  Show  how  to  construct  (a)  the  polar  plane  of  a  point  outside  of  a 
quadric,  (b)  the  pole  of  a  plane  which  cuts  the  quadric,  (c)  the  polar  plane  of 
a  point  within  a  quadric,  (d)  the  pole  of  a  plane  which  does  not  meet  the 
quadric. 

8.  Show  that  the  polar  plane  of  a  point  Pi  with  respect  to  a  sphere  is 
perpendicular  to  the  line  drawn  from  the  center  to  Pi- 

9.  Show  analytically  that  the  polar  plane  of  a  point  Pi  with  respect  to  a 
central  quadric  recedes  from  the  center  as  Pi  approaches  the  center,  and 
conversely. 

10.  Show  that  the  distances  from  two  points  to  the  center  of  a  sphere  are 
proportional  to  the  distances  of  each  of  these  points  from  the  polar  plane  of 
the  other. 

11.  Show  how  the  ideas  of  "polar  reciprocal  curves"  and  "polar  recip- 
rocation" with  respect  to  a  conic  may  be  generalized  to  "polar  reciprocal 
surfaces"  and  "polar  reciprocation"  with  respect  to  a  quadric. 

12.  What  is  the  polar  reciprocal  of  a  cone  or  cylinder  with  respect  to  a 
sphere  2  of  a  plane  curve  ? 

13.  Generalize  problem  7,  p.  320,  for  polar  reciprocation  with  respect  to 
a  quadric.  •' 

14.  Prove  that  the  distance  p  from  the  origin  to  the  plane  which  is  tangent 
to  the  ellipsoid       +  |-  +      ^  1  at  Pi  is  given  by  -^  =  :^  +  ^^  +  ^. 

a2         62         (.2  p2         (j4  ^4  c* 

15.  Prove  that  the  plane  Ax  +  By  -\-  Cz  -\-  D  =  0  is  tangent  to  the  ellipsoid 

X2         w2        2;2 

-  +  ^  +  1  =  lif  ^2^2  +  ^252  4.  C2c2  =  X>2. 

a2      o2      c2 


LINE  AND  QUADRIC  415 

16.  The  locus  of  the  point  of  intersection  of  three  mutually  perpendicular 
tangent  planes  to  an  ellipsoid  is  a  sphere  whose  radius  is  Va^  +  b^  -\^^. 

Hint.  From  problem  15  we  get  the  equations  of  three  tangent  planes.  Square  and 
add  these  equations,  making  use  of  the  conditions  that  the  planes  shall  be  mutually 
perpendicular. 

17.  Show  that  the  plane  Ax  +  By  +  Cz  +  D  =  0  is  tangent  to  the  parabo- 
loid —  ±y^  =  2czit  A^a^c  ±  B^b^c  =  2  CD. 

18.  Show  that  the  locus  of  the  point  of  intersection  of  three  mutually 
perpendicular  tangent  planes  to  a  paraboloid  is  a  plane. 

The  line  perpendicular  to  a  plane  which  is  tangent  to  a  surface 
at  the  point  of  tangency  is  called  the  normal  to  the  surface  at 
that  point. 

19.  Find  the  equation  of  the  normal  to  each  of  the  quadrics  at  a  point  Pi. 

20.  If  the  normal  to  an  ellipsoid  at  Pi  meets  the  principal  planes  in  ^,  B, 
and  C,  then  Pi^,  PiB,  and  PiC  are  in  a  constant  ratio. 

21.  Find  the  equation  of  the  cone  circumscribing  a  paraboloid  whose 
vertex  is  Pi  (xi,  ?/i,  Zi). 

22.  Find  the  equation  of  the  cylinder  circumscribing  an  ellipsoid  if  the 
direction  angles  of  the  elements  of  the  cylinder  are  a,  j8,  and  y. 

188.  Asymptotic  directions  and  cones.  If  the  coefficient  of  p"^  in  the 
equation  in  p  for  any  quadric  is  zero,  one  root  is  infinite  (Theorem  IV,  p.  15), 
and  the  line  meets  the  quadric  in  one  point  which  is  at  an  infinite  distance 
from  Pi.  The  direction  of  such  a  line  is  called  an  asymptotic  direction.  It  is 
evident  that  a  line  having  an  asijmptotic  direction  of  a  quadric  meets  the 
quadric  in  but  one  point  in  the  finite  part  of  space. 

It  is  easily  proved  that  the  coefficient  of  p2  is  formed  by  substituting 
cos  a,  cos  jS,  and  cos  y  for  x,  y,  and  z  in  the  terms  of  the  second  degree  in 
the  equation  of  the  quadric  (compare  the  footnote,  p.  236).  Hence  the 
direction  cosines  of  the  asymptotic  directions  of  the  non-degenerate  quadrics 

±^±^±?^.l,     ^^i^^  =  2c. 

respectively  satisfy  the  equations 

cos^a      cos2/3      COS27  cos^a      cos^/S 

(1)  "^-aT^-^^"^^^'    ~ar^~^=^' 


416  ANALYTIC  GEOMETRY 

By  considering  the  number  of  sets  of  real  numbers  satisfying  these 
equations  for  the  various  combinations  of  signs  we  obtain 

Theorem  m.  The  hyperboloids  and  the  hyperbolic  paraboloid  have  an 
infinite  number  of  asymptotic  directions,  the  elliptic  paraboloid  has  one,  and 
the  ellipsoid  has  none. 

The  lines  passing  through  a  given  point  Pi(xi,  yi,  Zi)  which  have  the 
asymptotic  directions  of  a  quadric  will,  in  general,  form  a  cone.  The 
equation  of  this  cone  for  the  hyperboloid  of  one  sheet 

a^2      y2      2;2 
^  '  a2       62       c2 

is  found  as  follows.      The   direction  cosines  of  an  asymptotic  direction 
satisfy  the  equation 

(3)  c^^c^_c<|.^„^  ^^^^^^^ 

If  the  equations  of  a  line  through  Pi  are 

(4)  x  =  Xi  +  p  cos  a,     2/  =  2/1  +  P  cos  /5,     z  —  Zi  +  p  cos  7, 

then 

X  —  Xi  ^      y  —  y\  z  —  Zi 

(5)  cos  a  = ,     cos  /3  = —  ,     cos  7  = . 

P  P  P 

Substituting  in  (3)  and  multiplying  by  p^,  we  get 

^  '  a2  52  c2 

as  the  condition  that  P  (x,  y,  z)  should  lie  on  a  line  through  Pi  which  has  an 
asymptotic  direction  of  (2).    Hence  (6)  is  the  equation  of  the  cone  whose 
vertex  is  Pi  and  whose  elements  have  the  asymptotic  directions  of  (2). 
That  (6)  is  really  the  equation  of  a  cone  is  verified  by  translating  the  origin  to  P^. 

In  general,  we  have  the 

Rule  to  find  the  equation  of  the  cone  of  asymptotic  directions  of  a  quadric 
whose  vertex  is  a  given  point. 

Set  the  coefficient  of  p^  in  the  equation  in  p  equal  to  zero,  and  substitute  the 
values  of  the  direction  cosines  derived  from  the  parametric  equations  of  the 
line. 

If  the  coefficients  of  p^  and  p  in  the  equation  in  p  are  both  zero,  then  both 
roots  are  infinite  *  (Theorem  IV,  p.  15)  and  the  line  is  called  an  asymptotic 
line. 

*  This  assumes  that  the  constant  term  is  not  zero.  If  the  constant  term  is  zero, 
Pi  lies  on  the  quadric,  and  when  the  coefficients  of  p^  and  p  are  both  zero,  any  number 
is  a  root  and  the  line  lies  entirely  on  the  quadric. 


LINE  AND  QUADRIC 


417 


Let  Pi  be  any  point  not  on  the  hyperboloid  (2)  and  let  us  seek  the  condi- 
tions that  a,  j8,  and  7  must  satisfy  if  the  line  (4)  is  an  asymptote. 
The  equation  in  p  for  the  hyperboloid  is 


cos^a      cos2/3      COS27 


\    a2 


62 


y+2( 


ccicosa      2/1  cos  ^      2^  COS  7  • 


+ 


\  a2       62 


+ 


62 


C2 


0. 


(7) 


If  (4)  is  an  asymptote,  then,  by  definition, 
COS27 


cos2a      cos2/3 
a2     "^     62 


c2 


=  0, 


Xi  COS  a      yi  cos  /S 
a2  62 


Zi COS  7 

C2 


=  0. 


These  are  therefore  the  conditions  which  a,  /3,  and  7  must  satisfy.  Equa- 
tions (7)  can  be  solved  for  cos  a  and  cos  /3 
in  terms  of  cos  7  and  there  will  be  two 
solutions  which  may  be  real  and  unequal, 
real  and  equal,  or  imaginary,  and  from  these 
we  can  determine  two  sets  of  numbers  to 
which  cos  a:,  cos/3,  and  cos  7  are  propor- 
tional. Hence  there  will  pass  through  Pi 
either  two  asymptotes,  one,  or  none. 

But  if  xi  =  yi  =  zi  =  0,  that  is,  if  Pi  is 
the  center  of  the  hyperboloid,  the  second  of 
equations  (7)  is  true  for  all  values  of  a,  jS, 
and  7 ;  and  as  the  first  of  equations  (7)  is 
identical  with  (3),  we  see  that  the  elements 
of  the  cone  of  asymptotic  directions  whose 
vertex  is  the  center  (0,  0,  0)  are  all  asymp- 
totic lines.  From  (6)  the  equation  of  this  cone,  which  is  called  the  asymptotic 
cone,  is  seen  to  be 

a2      62      c2 
Hence  we  have 


Theorem  IV.    The  equation  of  the  asymptotic  cone  of  the  hyperboloid  of  one 
sheet 


62 


a^      Ir      &• 


The  figure  shows  the  hyperboloid  (2)  in  outline  and  its  asymptotic  cone 
which  lies  entirely  within  the  surface.  As  the  hyperboloid  recedes  to  infinity 
it  approaches  closer  and  closer  to  its  asymptotic  cone  in  the  same  way  that 
an  hyperbola  approaches  its  asymptotes  (Theorem  IX,  p.  190). 


418 


ANALYTIC  GEOMETRY 


In  like  manner  we  may  prove  the  following  theorem, 

P 


Theorem  V.    The  equation  of  the  asymptotic  cone  of  the 

,2  ^,2  ^2 


hyperboloid  of  two  sheets 
hyperbolic  paraboloid 


2cz 


1  is 


O; 


The  latter  cone  degenerates  into  a  pair  of  intersecting  planes. 


PROBLEMS 

1.  Show  that  a  plane  perpendicular  to  the  axis  of  an  hyperbolic  parab- 
oloid intersects  the  surface  in  an  hyperbola  whose  asymptotes  form  the 
intersection  of  the  plane  with  the  asymptotic  cone. 

2.  Show  that  a  plane  passing  through  the  axis  of  an  hyperboloid  inter- 
sects the  surface  in  an  hyperbola  whose  asymptotes  form  the  intersection  of 
the  plane  with  the  asymptotic  cone. 

Hint.  Rotate  the  axes  about  the  axis  of  the  hyperboloid. 

3.  Show  that  the  asymptotic  directions  of  any  quadric  are  detern  ined 
by  the  locus  of  the  equation  obtained  by  setting  the  terms  of  the  second 
degree  equal  to  zero. 

4.  Show  that  a  plane  passing  through  the  center  and  a  generator  of  an 
hyperboloid  of  one  sheet  is  tangent  to  the  asymptotic  cone. 

6.  Show  that  any  plane  parallel  to  an  element  of  the  asymptotic  cone  of 
an  hyperboloid  intersects  the  hyperboloid  in  a  parabola. 

6.  Show  that  a  plane  tangent  to  the  asymptotic  cone  of  an  hyperboloid 
cuts  the  hyperboloid  in  two  parallel  lines. 

7.  Show  that  every  asymptotic  line  of  an  hyperboloid  is  parallel  to  an 
element  of  the  asymptotic  cone  and  lies  in  the  plane  tangent  to  the  cone 
along  that  element. 


LINK  AND  QUADRIC  419 

8.  By  means  of  problem  7  show  how  to  construct  the  asymptotic  lines 
of  an  hyperboloid  which  pass  through  any  point  Pi  other  than  the  center. 
Show  that  there  will  be  two,  one,  or  no  asymptotic  lines  through  Pi 
according  as  Pi  is  outside  of,  on,  or  inside  of  the  asymptotic  cone. 

x^       ?/2      z2  ^.2      yl      ^2 

9.  Show  that  the  hyperboloids =  1  and h  —  H —  =  1 

have  the  same  asymptotic  cone.    How  are  they  situated  relative  to  this  cone  ? 
10.  Show  that  two  asymptotes  of  an  hyperbolic  paraboloid  pass  through 
every  point  not  on  the  asymptotic  cone,  and  that  each  of  these  lines  is  par- 
allel to  one  of  the  planes  which  fortn  the  cone. 

189.  Centers.  A  point  Pi(xi,  yi,  Z\)  is  a  center  of  symmetry  of  a  quadric 
if  it  is  the  middle  point  of  every  chord  passing  through  it.  In  order  that  Pi 
shall  be  the  middle  point  of  a  chord,  the  roots  of  the  equation  in  p  must  be 
equal  numerically  with  opposite  signs,  and  hence  (Case  II,  p.  4)  the  coefficient 
of  p  must  be  zero.  The  coefficient  of  p  in  the  equation  in  p  for  the  general 
equation  of  the  second  degree  is  easily  seen  to  be 

(2  Axx  +  Fyi  +  Ezi  +  G)  cos  a 

+  (PiCi  +  2  %i  +  Dzx  +  H)  cos  iS  +  {Ex^  +  Dyi  +  2  Czi  +  I)  cos  7. 

This  is  zero  for  all  lines  passing  through  Pi,  that  is,  for  all  values  of 
cos  a,  cos  j3,  and  cos  7,  when  and  only  when  the  three  parentheses  are  zero. 
Setting  these 'parentheses  equal  to  zero  and  solving  for  Xi,  2/1,  and  Zi,  we  obtain 
the  coordinates  of  the  center. 

By  means  of  the  discussion  in  §  163,  p.  374,  we  see  that  a  quadric  may 
have  a  single  center,  that  there  may  be  no  center,  or  that  all  of  the  points 
of  a  line  or  of  a  plane  may  be  centers. 

190.  Diametral  planes.  The  locus  of  the  middle  points  of  a  system  of 
parallel  chords  of  a  quadric  is  found  to  be  a  plane  which  is  called  a  diametral 
planp. 

Consider  the  ellipsoid 

(1)  •  t  +  yl  +  ^  =  i 

^'  a'^      b^      c^ 

and  the  system  of  parallel  lines 

(2)  x  =  xi  +  p  cos  a,     y  =  yi  +  p  cos  jS,     z  =  zi  +  p  cos  7. 

These  equations  represent  a  system  of  parallel  lines  if  x^,yi,  and  Zj  are  arbitrary 
while  a,  /3,  and  y  are  constant. 

The  equation  in  p  for  (1)  is 

/cosVr      cos2/3      cos27\   ^         /Xicoscr      yiCos/3      2i_cos7\ 
^^       V    a2     "^     62     +     c2    r  V      a2      "^      62  c^     J^ 

^  V  a2  ^  62  ^  c2         / 


420  ANALYTIC  GEOMETRY 

If  Pi  is  the  middle  point  of  the  chord  of  (1)  formed  by  the  line  (2),  then 
the  roots  of  (3)  must  be  numerically  equal  with  opposite  signs ;  and  hence 
(Case  II,  p.  4) 

xi  cos  a      yicos^      ziC0S7  _  ^ 
^  ^  a^  Ifi  c^ 

is  the  condition  that  Pi  shall  be  the  middle  point  of  the  chord. 
But  (4)  is  the  condition  that  Pi  should  lie  in  the  plane 

X  cos  a      y  cos  /3      z  cos  7  _  ^ 
^2      +      62      +      c2      ""    ' 

and  this  is  therefore  the  equation  of  the  locus  of  the  middle  points  of  all 
chords  whose  direction  angles  are  or,  /3,  and  7. 

By  proceeding  in  this  manner  with  the  other  quadrics  we  obtain 
Theorem  VI.    The  equation  of  the  diametral  plane  bisecting  all  chords  whose 
direction  angles  are  a,  /3,  and  7  of  the 

x2     7/2     ^2  Oleosa      ycosB     zcosy 

central quadric         ±  -  ±  _  ±  -  =  Hs  ±  _^— i  — ^ +_ ^  =  O; 

x^     v^     ^  occosa      ycosB 

non-central  quadric      —  ±  --  =  2cz     is r-i-  =  c  COS  v. 


PROBLEMS 

1.  Determine  geometrically  the  number  of  centers  of  each  of  the  types  of 
quadrics  and  degenerate  quadrics. 

2.  Find  the  equation  of  the  diametral  plane  of  the  locus  of  the  general 
equation  of  the  second  degi-ee  bisecting  all  chords  whose  direction  angles  are 
a,  /3,  7.  From  the  form  of  the  equation  prove  that  the  plane  passes  through 
the  center  of  the  quadric  if  there  is  a  center. 

3.  Prove  that  every  plane  through  the  center  of  a  central  quadric  or 
parallel  to  the  axis  of  a  paraboloid  is  a  diametral  plane,  and  find  the  direc- 
tion cosines  of  the  chords  which  it  bisects. 

4.  The  line  of  intersection  of  two  diametral  planes  is  called  a  diameter. 
Show  that  a  central  quadric  has  three  diameters  such  that  the  plane  of  any 
two  bisects  all  chords  parallel  to  the  third.  Such  lines  are  called  conjugate 
diameters,  and  the  plane  of  any  two  is  said  to  be  conjugate  to  the  third. 

5.  Find  the  equation  of  the  plane  which  bisects  all  chords  of  (a)  a  cen- 
tral quadric,  (b)  a  paraboloid,  which  are. parallel  to  the  diameter  passing 
through  a  point  Pi  on  the  quadric. 

6.  The  planes  tangent  to  a  quadric  at  the  extremities  of  a  diameter  are 
parallel  to  the  conjugate  diametral  plane. 


LINE  AND  QUADRIC  421 

7.  The  sum  of  the  squares  of  the  projections  of  three  conjugate  semi- 
diameters  of  an  ellipsoid  on  each  of  the  axes  of  the  ellipsoid  is  constant. 

Hint.  Let  P^,  P^,  and  P^  be  the  extremities  of  three  conjugate  diameters.  Find  tlie 
conditions  that  tliese  points  are  on  the  ellipsoid  and  that  any  two  are  on  the  plane 
conjugate  to  the  diameter  through  the  third.    Then  show  that 

a^   b^    c^     a'    b  ^    c'  a  '  b  ^  c 

are  the  direction  cosines  of  three  mutually  perpendicular  lines,  and  that  if  these  lines 
be  chosen  as  axes,  then 

a^  a^  a^     b^   b '  b  '  c  '  c  '  c 

are  also  the  direction  cosines  of  three  lines.    Then  apply  Theorem  III,  p.  330. 

8.  By  means  of  problem  7  show  that  the  sum  of  the  squares  of  three 
conjugate  semi-diameters  of  an  ellipsoid  is  equal  to  a^  +  6^  4.  c2. 


IP 


INDEX 


Abscissa,  24 

Absolute  invariant,  270 

Algebraic  equation,  17  ;  curve,  72 

Anchor  ring,  389 

Arbitrary  constant,  1 

Auxiliary  circle,  206 


Euclidean  transformation,  281 
External  angle,  121 


Fixed  point,  285 

Focal  radii  of  conies,  193, 

Four-leaved  rose,  263 


194 


Cardioid,  158,  253,  302 

Center  of  similitude  of  circles,  296 

Central  conic,  183 ;  quadric,  399 

Cissoid  of  Diodes,  253,  263,  299 

Complete  quadrilateral,  123 

Conchoid  of  Nicomedes,  251 

Condition  for  tangency,  230 

Confocal  conies,  203 

Congruent  figures,  281 

Conicoid,  397 

Conjugate  diameters  of  quadrics,  420 

Cubical  parabola,  72 

Curtate  cycloid,  259 

Cycloid,  256 

Degenerate  ellipse,  195;  hyperbola, 
195;  parabola,  196;  quadric,  399 

Direction  cosines  of  a  line,  123,  330, 
364 

Director  circle,  261 

Discriminant  of  the  equation  of  a 
circle,  131 ;  of  the  general  equation 
of  the  second  degree,  265;  of  a 
quadratic,  2 


Graph  of  an  equation,  83 

nomothetic  figures,  291  * 

Hyperbolic  spiral,  249 
Hypocycloid,  259 ;  of  four  cusps,  257 

Intercepts,  73 

Internal  angle,  121 

Invariant,  270 ;  line,  288 ;  point,  285 

Involute  of  a  circle,  259 

Latus  rectum,  181 

Lemniscate   of  Bernoulli,  153,  248, 

262,  300 
Lima^on  of  Pascal,  253,  262,  302 
Limiting  points  of  a  system  of  circles, 

144 

Normal  to  a  curve,  210;  length  of, 
214 ;  to  a  surface,  415 

Ordinate,  24 

Orthogonal  circles,  143;  systems  of 
circles,  308 


Epicycloid,  259  Parabolic  spiral,  263 

Equations  of  a  transformation,  281         Parameter,  1 

423 


424 


INDEX 


Peaucellier's  Inversor,  309 
Point-circle,  131 

Polar  reciprocal  curves,  313,  317 
Prolate  cycloid,  259 

Radian,  19 

Radical  axis,  137,  382 

Reciprocal  spiral,  249 


Subnormal,  214 
Subtangent,  214 

Torus,  389 

Traces  of  a  surface,  346 
Transcendental  equation,  17 
Transformation,  281 
Trisectrix  of  Maclaurin,  303 


Self-conjugate  triangles,  319 
Semicubical  parabola,  209 
Spiral  of  Archimedes,  249 
Strophoid,  262,  263,  301 


Vertex  of  a  conic,  174,  175 
Witch  of  Agnesi,  250 


^     OF   THE 

UNIVERSITY 


IP 


14  DAY  USE 

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This  book  is  due  on  the  last  date  stamped  below  or 
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